Rd Sharma 2020 Solutions for Class 10 Maths Chapter 10 Trigonometric Ratios are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios are extremely popular among Class 10 students for Maths Trigonometric Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 Book of Class 10 Maths Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 Solutions. All Rd Sharma 2020 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 10.23:

#### Question 1:

In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.

(i) $\mathrm{sin}A=\frac{2}{3}$

(ii) $\mathrm{cos}A=\frac{4}{5}$

(iii) tan θ = 11

(iv) $\mathrm{sin}\theta =\frac{11}{15}$

(v) $\mathrm{tan}\mathrm{\alpha}=\frac{5}{12}$

(vi) $\mathrm{sin}\theta =\frac{\sqrt{3}}{2}$

(vii) $\mathrm{cos}\theta =\frac{7}{25}$

(viii) $\mathrm{tan}\theta =\frac{8}{15}$

(ix) $\mathrm{cot}\theta =\frac{12}{5}$

(x) $\mathrm{sec}\theta =\frac{13}{5}$

(xi) $\mathrm{cosec}\mathrm{\theta}=\sqrt{10}$

(xii) $\mathrm{cos}\theta =\frac{12}{15}$

#### Answer:

(i) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = 2 and

Hypotenuse = 3

Therefore, by Pythagoras theorem,

Now we substitute the value of perpendicular side (*BC*) and hypotenuse (*AC*) and get the base side (*AB*)

Therefore,

Hence, Base =

Now,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(ii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 4 and

Hypotenuse = 5

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (*AB*) and hypotenuse (*AC*) and get the perpendicular side (*BC*)

Hence, Perpendicular side = 3

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(iii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 1 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (*AB*) and the perpendicular side (*BC*) and get hypotenuse (*AC*)

Hence, Hypotenuse =

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(iv) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = 11 and

Hypotenuse = 15

Therefore,

By Pythagoras theorem,

Now we substitute the value of perpendicular side (*BC*) and hypotenuse(*AC*) and get the base side (*AB*)

Hence, Base =

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(v) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 12 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (*AB*) and the perpendicular side (*BC*) and get hypotenuse (*AC*)

Hence, Hypotenuse = 13

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(vi) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = and

Hypotenuse = 2

Therefore,

By Pythagoras theorem,

Now we substitute the value of perpendicular side (*BC*) and hypotenuse(*AC*) and get the base side (*AB*)

Hence, Base = 1

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(vii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 7 and

Hypotenuse = 25

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (*AB*) and hypotenuse (*AC*) and get the perpendicular side (*BC*)

Hence, Perpendicular side = 24

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(viii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 15 and

Perpendicular side = 8

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (*AB*) and the perpendicular side (*BC*) and get hypotenuse (*AC*)

Hence, Hypotenuse = 17

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(ix) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 12 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

*AB*) and the perpendicular side (*BC*) and get hypotenuse (*AC*)

Hence, Hypotenuse = 13

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(x) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 5 and

Hypotenuse = 13

Therefore,

By Pythagoras theorem,

Now we substitute the value of base side (*AB*) and hypotenuse (*AC*) and get the perpendicular side (*BC*)

Hence, Perpendicular side = 12

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(xi) Given:

…… (1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = 1 and

Hypotenuse =

Therefore,

By Pythagoras theorem,

Now we substitute the value of perpendicular side (*BC*) and hypotenuse (*AC*) and get the base side (*AB*)

Hence, Base side = 3

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

(xii) Given: ……(1)

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Base = 12 and

Hypotenuse = 15

Therefore,

By Pythagoras theorem,

*AB*) and hypotenuse (*AC*) and get the perpendicular side (*BC*)

Hence, Perpendicular side = 9

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

Now,

Therefore,

#### Page No 10.23:

#### Question 2:

In a ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine

(i) sin A, cos A

(ii) sin C, cos C

#### Answer:

(i) The given triangle is below:-

Given: In ,

To Find:

In this problem, Hypotenuse side is unknown

Hence we first find Hypotenuse side by Pythagoras theorem

By Pythagoras theorem,

We get,

By definition,

By definition,

Answer:

(ii) The given triangle is below:

Given: In ΔABC,

To Find:

In this problem, Hypotenuse side is unknown

Hence we first find Hypotenuse side by Pythagoras theorem

By Pythagoras theorem,

We get,

By definition,

By definition,

Answer:

#### Page No 10.23:

#### Question 3:

In the given figure, find tan P and cot R. Is tan P = cot R?

#### Answer:

The given figure is below:

To Find:

In the given right angled ΔPQR, length of side QR is unknown.

Therefore, to find length of side QR we use Pythagoras Theorem

Hence, by applying Pythagoras theorem in ΔPQR,

We get,

Now, we substitute the length of given side PR and PQ in the above equation

By definition, we know that

… (1)

Also, by definition, we know that

… (2)

Comparing equation (1) and (2) , we come to know that R.H.S of both the equation are equal

Therefore, L.H.S of both the equation are also equal

Answer:

#### Page No 10.24:

#### Question 4:

If $\mathrm{sin}A=\frac{9}{41}$, compute cos *A* and tan *A*.

#### Answer:

Given: ……(1)

To Find:

By definition,

…... (2)

By Comparing (1) and (2)

We get,

Perpendicular side = 9 and

Hypotenuse = 41

Now using the perpendicular side and hypotenuse we can construct as shown below

Length of side AB is unknown in right angled ,

To find length of side AB, we use Pythagoras theorem.

Therefore, by applying Pythagoras theorem in ,

We get,

Hence, length of side AB = 40

Now,

By definition,

Now,

By definition,

Answer: and

#### Page No 10.24:

#### Question 5:

Given 15 cot *A* = 8, find sin *A* and sec *A*.

#### Answer:

Given: 15 = 8

To Find:

Since

By taking 15 on R.H.S

We get,

By definition,

Hence,

Comparing equation (1) and (2)

We get,

= 8

= 15

can be drawn as shown below using above information

Hypotenuse side AC is unknown.

Therefore, we find side AC of by Pythagoras theorem.

So, by applying Pythagoras theorem to

We get,

Therefore, Hypotenuse = 17

Now by definition,

Substituting values of sides from the above figure

By definition,

Hence,

Answer: and

#### Page No 10.24:

#### Question 6:

In ∆PQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R.

#### Answer:

Given:

is right angled at vertex *Q*.

To find:

Given is as shown below

Hypotenuse side *PR* is unknown.

Therefore, we find side *PR* of by Pythagoras theorem

By applying Pythagoras theorem to

We get,

Hence, Hypotenuse = 5

Now by definition,

Substituting values of sides from the above figure

Now by definition,

Substituting values of sides from the above figure

By definition,

By definition,

Substituting values of sides from the above figure

Answer: , , and

#### Page No 10.24:

#### Question 7:

If $\mathrm{cot}\mathrm{\theta}=\frac{7}{8}$, evaluate :

(i) $\frac{\left(1+\mathrm{sin}\theta \right)\left(1-\mathrm{sin}\theta \right)}{\left(1+\mathrm{cos}\theta \right)\left(1-\mathrm{cos}\theta \right)}$

(ii) cot^{2} θ

#### Answer:

(i) Given:

To evaluate:

…… (1)

We know the following formula

By applying the above formula in the numerator of equation (1) ,

We get,

… (Where *a* = 1 and *b* = )

…… (2)

Similarly,

By applying formula in the denominator of equation (1) ,

We get,

… (Where *a* = 1 and *b* = )

…… (3)

Substituting the value of numerator and denominator of equation (1) ,from equation (2) and(3)

Therefore,

…… (4)

Since,

Therefore,

Putting the value of and in Equation (4)

We get,

We know that,

Since,

Therefore,

Answer:

(ii) Given:

To evaluate:

Squaring on both sides,

We get,

Answer:

#### Page No 10.24:

#### Question 8:

If 3 cot A = 4, check whether $\frac{1-{\mathrm{tan}}^{2}A}{1+{\mathrm{tan}}^{2}A}={\mathrm{cos}}^{2}A-{\mathrm{sin}}^{2}A$ or not.

#### Answer:

Given:

To check whether or not

Dividing by 3 on both sides,

We get,

…… (1)

By definition,

Therefore,

…… (2)

Comparing Equation (1) and (2)

We get,

= 4

= 3

Hence, is as shown in figure below

In, Hypotenuse is unknown

Hence, It can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem in

We get

Hence, Hypotenuse = 5

To check whether or not

We get the values of

By definition,

Substituting the value from Equation (1)

We get,

….… (3)

Now by definition,

…… (4)

Now by definition,

…… (5)

Now we first take L.H.S of Equation

Substituting value of from equation (3)

We get,

Taking L.C.M on both numerator and denominator

We get,

…… (6)

Now we take R.H.S of Equation

Substituting value of and from equation (4) and (5)

We get,

…… (7)

Comparing Equation (6) and (7)

We get,

Answer: Yes

#### Page No 10.24:

#### Question 9:

If $\mathrm{tan}\theta =\frac{a}{b}$, find the value of $\frac{\mathrm{cos}\theta +\mathrm{sin}\theta}{\mathrm{cos}\theta -\mathrm{sin}\theta}$.

#### Answer:

Given:

…… (1)

Now, we know that

Therefore equation (1) becomes as follows

Now, by applying invertendo

We get,

Now, by applying Compenendo-dividendo

We get,

Therefore,

#### Page No 10.24:

#### Question 10:

If 3 tan θ = 4, find the value of $\frac{4\mathrm{cos}\theta -\mathrm{sin}\theta}{2\mathrm{cos}\theta +\mathrm{sin}\theta}$.

#### Answer:

Given:

Therefore,

…… (1)

Now, we know that

Therefore equation (1) becomes

…… (2)

Now, by applying Invertendo to equation (2)

We get,

…… (3)

Now, multiplying by 4 on both sides

We get,

Therefore

Now by applying dividendo in above equation

We get,

…… (4)

Now, multiplying by 2 on both sides of equation (3)

We get,

Therefore

Now by applying componendo in above equation

We get,

…… (5)

Now, by dividing equation (4) by equation (5)

We get,

Therefore,

Therefore, on L.H.S cancels and we get,

Therefore,

Hence,

#### Page No 10.24:

#### Question 11:

If 3 cot θ = 2, find the value of $\frac{4\mathrm{sin}\theta -3\mathrm{cos}\theta}{2\mathrm{sin}\theta +6\mathrm{cos}\theta}$.

#### Answer:

Given:

Therefore,

…… (1)

Now, we know that

Therefore equation (1) becomes

…… (2)

Now, by applying Invertendo to equation (2)

We get,

…… (3)

Now, multiplying by on both sides

We get,

Therefore, 3 cancels out on R.H.S and

We get,

Now by applying dividendo in above equation

We get,

…… (4)

Now, multiplying by on both sides of equation (3)

We get,

Therefore, 2 cancels out on R.H.S and

We get,

Now by applying componendo in above equation

We get,

…… (5)

Now, by dividing equation (4) by equation (5)

We get,

Therefore,

Therefore, on L.H.S cancels out and we get,

Now, by taking 2 in the numerator of L.H.S on the R.H.S

We get,

Therefore, 2 cancels out on R.H.S. and

We get,

Hence,

#### Page No 10.24:

#### Question 12:

If $\mathrm{tan}\theta =\frac{a}{b}$, prove that $\frac{a\mathrm{sin}\theta +b\mathrm{cos}\theta}{a\mathrm{sin}\theta +b\mathrm{cos}\theta}=\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}$.

#### Answer:

Given:

…… (1)

Now, we know that

Therefore equation (1) becomes

…… (2)

Now, multiplying by on both sides of equation (2)

We get,

Therefore,

…... (3)

Now by applying dividendo in above equation (3)

We get,

…… (4)

Now by applying componendo in equation (3)

We get,

…… (5)

Now, by dividing equation (4) by equation (5)

We get,

Therefore,

Therefore, and cancels on L.H.S and R.H.S respectively and we get,

Hence, it is proved that

#### Page No 10.24:

#### Question 13:

If $\mathrm{sec}\theta =\frac{13}{5}$, show that $\frac{2\mathrm{sin}\theta -3\mathrm{cos}\theta}{4\mathrm{sin}\theta -9\mathrm{cos}\theta}=3$.

#### Answer:

Given:

To show that

Now, we know that

Therefore,

Therefore,

…… (1)

Now, we know that

…… (2)

Now, by comparing equation (1) and (2)

We get,

= 5

And

Hypotenuse = 13

Therefore from above figure

Base side

Hypotenuse

Side *AB* is unknown, It can be determined by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

Therefore by substituting the values of known sides

We get,

Therefore,

Therefore,

…… (3)

Now, we know that

Now from figure (a)

We get,

Therefore,

…… (4)

Now L.H.S. of the equation to be proved is as follows

Substituting the value of and from equation (1) and (4) respectively

We get,

Therefore,

Hence proved that,

#### Page No 10.24:

#### Question 14:

If $\mathrm{cos}\theta =\frac{12}{13}$, show that sin θ (1 − tan θ)$=\frac{35}{156}$.

#### Answer:

Given: ……(1)

To show that

Now, we know that …… (2)

Therefore, by comparing equation (1) and (2)

We get,

= 12

And

Hypotenuse = 13

Therefore from above figure

Base side

Hypotenuse

Side AB is unknown and it can be determined by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

Therefore by substituting the values of known sides

We get,

Therefore,

Therefore,

…… (3)

Now, we know that

Now from figure (*a*)

We get,

Therefore,

…… (4)

Now, we know that

Now from figure (a)

We get,

Therefore,

…… (5)

Now L.H.S. of the equation to be proved is as follows

…… (6)

Substituting the value of and from equation (4) and (5) respectively

We get,

Taking L.C.M inside the bracket

We get,

Therefore,

Now, by opening the bracket and simplifying

We get,

…… (7)

From equation (6) and (7) , it can be shown that

#### Page No 10.24:

#### Question 15:

If $\mathrm{cot}\theta =\frac{1}{\sqrt{3}}$, show that $\frac{1-{\mathrm{cos}}^{2}\theta}{2-{\mathrm{sin}}^{2}\theta}=\frac{3}{5}$.

#### Answer:

Given: ……(1)

To show that

Now, we know that

Since

Therefore,

Therefore,

…… (2)

Comparing Equation (1) and (2)

We get,

= 1

Therefore, Triangle representing angle is as shown below

Hypotenuse *AC* is unknown and it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

Therefore by substituting the values of known sides

We get,

Therefore,

Therefore,

…… (3)

Now, we know that

Now from figure (*a*)

We get,

Therefore from figure (a) and equation (3) ,

…… (4)

Now, we know that

Now from figure (*a*)

We get,

Therefore from figure (*a*) and equation (3) ,

…… (5)

Now, L.H.S of the equation to be proved is as follows

Substituting the value of and from equation (4) and (5)

We get,

Now by taking L.C.M. in numerator as well as denominator

We get,

Therefore,

Therefore,

Therefore,

Hence proved that

#### Page No 10.24:

#### Question 16:

If $\mathrm{tan}\theta =\frac{1}{\sqrt{7}}$, show that $\frac{{\mathrm{cosec}}^{2}\theta -{\mathrm{sec}}^{2}\theta}{{\mathrm{cosec}}^{2}\theta +{\mathrm{sec}}^{2}\theta}=\frac{3}{4}$

#### Answer:

Given: ……(1)

To show that

Now, we know that

Since ……(2)

Therefore,

Comparing Equation (1) and (2)

We get,

Therefore, Triangle representing angle is as shown below

Hypotenuse *AC* is unknown and it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

Therefore by substituting the values of known sides

We get,

Therefore,

Therefore,

…… (3)

Now, we know that

Now from figure (a)

We get,

Therefore from figure (a) and equation (3) ,

…… (4)

Now, we know that

Therefore, from equation (4)

We get,

Therefore,

…… (5)

Now, we know that

Now from figure (a)

We get,

Therefore from figure (a) and equation (3) ,

…… (6)

Now, we know that

Therefore, from equation (6)

We get,

Therefore,

…… (7)

Now, *L.H.S* of the equation to be proved is as follows

Substituting the value of and from equation (6) and (7)

We get,

Now by taking L.C.M. in numerator as well as denominator

We get,

Therefore,

Therefore,

Therefore,

Hence proved that

#### Page No 10.24:

#### Question 17:

If $\mathrm{sec}\theta =\frac{5}{4}$, find the value of $\frac{\mathrm{sin}\theta -2\mathrm{cos}\theta}{\mathrm{tan}\theta -\mathrm{cot}\theta}$.

#### Answer:

Given: ……(1)

To find the value of

Now we know that

Therefore,

Therefore from equation (1)

…… (2)

Also, we know that ${\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1$

Therefore,

Substituting the value of from equation (2)

We get,

Therefore

…… (3)

Also, we know that $se{c}^{2}\theta =1+{\mathrm{tan}}^{2}\theta $.

Therefore,

${\mathrm{tan}}^{2}\theta =se{c}^{2}\theta -1$

Therefore

${\mathrm{tan}}^{2}\theta ={\left(\frac{5}{4}\right)}^{2}-1\phantom{\rule{0ex}{0ex}}=\frac{25}{16}-1\phantom{\rule{0ex}{0ex}}=\frac{9}{16}$

Therefore,

$\mathrm{tan}\theta =\sqrt{\frac{9}{16}}\phantom{\rule{0ex}{0ex}}=\frac{3}{4}$

Therefore,

$\mathrm{tan}\theta =\frac{3}{4}$ …… (4)

Also $cot\theta =\frac{1}{\mathrm{tan}\theta}$

Therefore, from equation (4)

We get,

$cot\theta =\frac{1}{{\displaystyle \frac{3}{4}}}$

$cot\theta =\frac{4}{3}$…… (5)

Substituting the value of,,and from equation (2) (3) (4) and (5) respectively in the expression below

We get,

$\frac{\mathrm{sin}\theta -2\mathrm{cos}\theta}{\mathrm{tan}\theta -\mathrm{cot}\theta}=\frac{{\displaystyle \frac{3}{5}}-2\left({\displaystyle \frac{4}{5}}\right)}{{\displaystyle \frac{3}{4}}-{\displaystyle \frac{4}{3}}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{3}{5}}-{\displaystyle \frac{8}{5}}}{{\displaystyle \frac{\left(3\times 3\right)-\left(4\times 4\right)}{4\times 3}}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{3}{5}}-{\displaystyle \frac{8}{5}}}{{\displaystyle \frac{\left(3\times 3\right)-\left(4\times 4\right)}{4\times 3}}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{3-8}{5}}}{{\displaystyle \frac{9-16}{4\times 3}}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{-5}{5}}}{{\displaystyle \frac{-7}{12}}}\phantom{\rule{0ex}{0ex}}=\frac{12}{7}$

Therefore,

#### Page No 10.24:

#### Question 18:

If $\mathrm{tan}\theta =\frac{12}{13}$, find the value of $\frac{2\mathrm{sin}\theta \mathrm{cos}\theta}{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta}$

#### Answer:

Given: ……(1)

To find the value of

Now, we know the following trigonometric identity

Therefore, by substituting the value of from equation (1) ,

We get,

By taking L.C.M. on the R.H.S,

We get,

Therefore

Therefore

…… (2)

Now, we know that

Therefore,

Therefore

…… (3)

Now, we know the following trigonometric identity

Therefore,

Now by substituting the value of from equation (3)

We get,

Therefore, by taking L.C.M on R.H.S

We get,

Now, by taking square root on both sides

We get,

Therefore,

…… (4)

Substituting the value of and from equation (3) and (4) respectively in the expression below

Therefore,

Therefore,

#### Page No 10.25:

#### Question 19:

If $\mathrm{cos}\theta =\frac{3}{5}$, find the value of $\frac{\mathrm{sin}\theta -{\displaystyle \frac{1}{\mathrm{tan}\theta}}}{2\mathrm{tan}\theta}$.

#### Answer:

Given: ……(1)

To find the value of

Now, we know the following trigonometric identity

Therefore, by substituting the value of from equation (1) ,

We get,

Therefore,

Therefore by taking square root on both sides

We get,

Therefore,

…… (2)

Now, we know that

Therefore by substituting the value of and from equation (2) and (1) respectively

We get,

…… (4)

Now, by substituting the value of and from equation (2) and (4) respectively in the expression below

We get,

Therefore,

Therefore,

#### Page No 10.25:

#### Question 20:

If $\mathrm{sin}\theta =\frac{3}{5}$, evaluate $\frac{\mathrm{cos}\theta -{\displaystyle \frac{1}{\mathrm{tan}\theta}}}{2\mathrm{cot}\theta}$.

#### Answer:

Given: ……(1)

To find the value of

Now, we know the following trigonometric identity

Therefore, by substituting the value of from equation (1) ,

We get,

Therefore,

Now by taking L.C.M

We get,

Therefore by taking square root on both sides

We get,

Therefore,

…… (2)

Now, we know that

Therefore by substituting the value of and from equation (1) and (2) respectively

We get,

…… (3)

Also, we know that

Therefore from equation (4) ,

We get,

Therefore,

…… (4)

Now, by substituting the value of, and from equation (2) , (3) and (4) respectively in the expression below

We get,

Therefore,

#### Page No 10.25:

#### Question 21:

If $\mathrm{tan}\theta =\frac{24}{7}$, find that sin θ + cos θ.

#### Answer:

Given:

…… (1)

To find:

Now we know is defined as follows

…… (2)

Now by comparing equation (1) and (2)

We get

= 24

= 7

Therefore triangle representing angle is as shown below

Side *AC* is unknown and can be found using Pythagoras theorem

Therefore,

Now by substituting the value of known sides from figure

We get,

Now by taking square root on both sides

We get,

Therefore Hypotenuse side AC = 25 …… (3)

Now we know, is defined as follows

Therefore from figure (a) and equation (3)

We get,

…… (4)

Now we know, is defined as follows

Therefore from figure (a) and equation (3)

We get,

…… (5)

Now we need to find the value of expression

Therefore by substituting the value of andfrom equation (4) and (5) respectively, we get,

Hence

#### Page No 10.25:

#### Question 22:

If $\mathrm{sin}\theta =\frac{a}{b}$, find sec θ + tan θ in terms of *a* and *b*.

#### Answer:

Given:

…… (1)

To find:

Now we know, is defined as follows

…… (2)

Now by comparing (1) and (2)

We get,

= a

and

Hypotenuse = b

Therefore triangle representing angle is as shown below

Here side *BC* is unknown

Now we find side *BC* by applying Pythagoras theorem to right angled

Therefore,

Now by substituting the value of sides *AB* and *AC* from figure (a)

We get,

Therefore,

Now by taking square root on both sides

We get,

Therefore,

Base side …… (3)

Now we know, is defined as follows

Therefore from figure (*a*) and equation (3)

We get,

…… (4)

Now we know,

Therefore,

Therefore,

…… (5)

Now we know,

Now by substituting the values from equation (1) and (3)

We get,

Therefore,

…… (6)

Now we need to find

Now by substituting the value of and from equation (5) and (6) respectively

We get,

…… (7)

Now we have the following formula which says

Therefore by applying above formula in equation (7)

We get,

Now by substituting in above expression

We get,

Now present in the numerator as well as denominator of above expression gets cancels and we get,

Square root is present in the numerator as well as denominator of above expression

Therefore we can place both numerator as well as denominator under a common square root sign

Therefore,

#### Page No 10.25:

#### Question 23:

If 8 tan A = 15, find sin A − cos A.

#### Answer:

Given:

Therefore,

…… (1)

To find:

Now we know is defined as follows

…… (2)

Now by comparing equation (1) and (2)

We get

= 15

= 8

Therefore triangle representing angle A is as shown below

Side *AC* is unknown and can be found using Pythagoras theorem

Therefore,

Now by substituting the value of known sides from figure (a)

We get,

Now by taking square root on both sides

We get,

Therefore Hypotenuse side AC = 17 …… (3)

Now we know, is defined as follows

Therefore from figure (a) and equation (3)

We get,

…… (4)

Now we know, is defined as follows

Therefore from figure (a) and equation (3)

We get,

…… (5)

Now we need to find the value of expression

Therefore by substituting the value of andfrom equation (4) and (5) respectively, we get,

Hence

#### Page No 10.25:

#### Question 24:

If $\mathrm{tan}\theta =\frac{20}{21}$, show that $\frac{1-\mathrm{sin}\theta +\mathrm{cos}\theta}{1+\mathrm{sin}\theta +\mathrm{cos}\theta}=\frac{3}{7}$.

#### Answer:

Given:

…… (1)

To show that:

Now we know is defined as follows

…… (2)

Now by comparing equation (1) and (2)

We get

= 20

= 21

Therefore triangle representing angle is as shown below

Side AC is unknown and can be found using Pythagoras theorem

Therefore,

Now by substituting the value of known sides from figure (a)

We get,

Now by taking square root on both sides

We get,

Therefore Hypotenuse side AC = 29 …… (3)

Now we know, is defined as follows

Therefore from figure (a) and equation (3)

We get,

…… (4)

Now we know, is defined as follows

Therefore from figure (a) and equation (3)

We get,

…… (5)

Now we need to find the value of expression

Therefore by substituting the value of andfrom equation (4) and (5) respectively, we get,

Now by taking L.C.M on R.H.S of above equation

We get

Now as 10 is present in numerator as well as denominator of R.H.S of above equation, it gets cancelled and we get

Hence

#### Page No 10.25:

#### Question 25:

If cosec A = 2, find the value of $\frac{1}{\mathrm{tan}A}+\frac{\mathrm{sin}A}{1+\mathrm{cos}A}$.

#### Answer:

Given:

…… (1)

To find:

Now we know is defined as below

Therefore,

Now by substituting the value of from equation (1)

We get,

…… (2)

Now by substituting the value of in the following identity of trigonometry

We get,

Now by taking L.C.M we get

Now by taking square root on both sides

We get,

Therefore,

…… (3)

Now is defined as follows

Now by substituting the value of and from equation (2) and (3) respectively we get,

Therefore,

…… (4)

Now by substituting the value of, and from equation (2) , (3) and (4) respectively we get,

Now by taking L.C.M we get

Now 2 gets cancelled and we get

Now by taking L.C.M, we get,

Now by opening the brackets in the numerator

We get,

Therefore,

Now by taking 2 common

We get,

Now as is present in both numerator as well as denominator, it gets cancelled

Therefore,

#### Page No 10.25:

#### Question 26:

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

#### Answer:

Given:

…… (1)

To show:

is as shown in figure below

Now since ……from (1)

Therefore

Now observe that denominator of above equality is same that is *AB*

Hence only when

Therefore …… (2)

We know that when two sides of a triangle are equal, then angle opposite to the sides are also equal.

Therefore from equation (2)

We can say that

Angle opposite to side *AC* = Angle opposite to side *BC*

Therefore,

Hence,

#### Page No 10.25:

#### Question 27:

In a ∆ABC, right angled at A, if $\mathrm{tan}\mathrm{C}=\sqrt{3}$, find the value of sin B cos C + cos B sin C.

#### Answer:

Given:

To find:

The givenis as shown in figure below

Side *BC* is unknown and can be found using Pythagoras theorem

Therefore,

Now by substituting the value of known sides from figure (*a*)

We get,

Now by taking square root on both sides

We get,

Therefore Hypotenuse side *BC* = 2 …… (1)

Now

Therefore,

Now by substituting the values from equation (1) and figure (*a*)

We get,

…… (2)

Now

Therefore,

Now by substituting the values from equation (1) and figure (a)

We get,

…… (3)

Now

Therefore,

Now by substituting the values from equation (1) and figure (*a*)

We get,

…… (4)

Now by definition,

Therefore,

Now by substituting the value of and from equation (4) and given data respectively

We get,

Now gets cancelled as it is present in both numerator and denominator

Therefore,

…… (5)

Now by substituting the value of and from equation (2) , (3) , (4) and (5) respectively in

We get,

#### Page No 10.25:

#### Question 28:

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) $\mathrm{sec}A=\frac{12}{5}$ for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) $\mathrm{sin}\theta =\frac{4}{3}$for some angle θ.

#### Answer:

(i) In, is acute an angle

Therefore,

Minimum value of is 0° and

Maximum value of is 90°

We know that and

tan90° = ∞

Therefore the statement that;

“The value of is always less than1” is false

(ii)

In and, is acute angle

Therefore,

Minimum value of is 0°and

Maximum value of is

We know that cos0° = 1 and

cos90° = 0

Now,

Therefore minimum value of is …… (1)

Now,

Therefore maximum value of is …… (2)

Now consider the given value

Here,

This value 2.4 lies in between 1 and

Now from equation (1) and (2) , we can say that the value lies in between minimum value of (that is 1) and maximum value of (that is)

Hence, , for some value of angle A is true

(iii) Cosecant of angle A is defined as

Also, is defined as

Therefore,

…… (1)

And

is defined as …… (2)

Therefore from equation (1) and (2) , it is clear that and (that is cosecant of angle A) are two different trigonometric angles

Hence, is the abbreviation used for cosecant of angle A is False

(iv) cot A is a trigonometric ratio which means cotangent of angle A

Hence, is the product of cot and A is False

(v)

The value

In, is acute an angle

Therefore,

Minimum value of is 0° and

Maximum value of is 90°

We know that and

sin90° = 1

Therefore the value of should lie between 0 and 1 and must not exceed 1

Hence the given value for (that is) is not possible

Therefore, , for some angle = False

#### Page No 10.25:

#### Question 29:

If $\mathrm{sin}\theta =\frac{12}{13}$, find the value of $\frac{{\mathrm{sin}}^{2}\theta -{\mathrm{cos}}^{2}\theta}{2\mathrm{sin}\theta \mathrm{cos}\theta}\times \frac{1}{{\mathrm{tan}}^{2}\theta}$.

#### Answer:

Given: ……(1)

To Find: The value of expression

Now, we know that

…… (2)

Now when we compare equation (1) and (2)

We get,

= 12

And

Hypotenuse = 13

Therefore, Triangle representing angle is as shown below

Base side *BC* is unknown and it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

Therefore by substituting the values of known sides

We get,

Therefore,

Therefore,

…… (3)

Now, we know that

Now from figure (a)

We get,

Therefore from figure (*a*) and equation (3) ,

…… (4)

Now we know that,

Therefore, substituting the value of and from equation (1) and (4)

We get,

Therefore 13 gets cancelled and we get

…… (5)

Now we substitute the value of , and from equation (1) , (4) and (5) respectively in the expression below

Therefore,

We get,

Therefore by further simplifying we get,

Now 169 gets cancelled and gets reduced to

Therefore

Therefore the value of is

That is

#### Page No 10.25:

#### Question 30:

If $\mathrm{cos}\theta =\frac{5}{13}$, find the value of $\frac{{\mathrm{sin}}^{2}\theta -{\mathrm{cos}}^{2}\theta}{2\mathrm{sin}\theta \mathrm{cos}\theta}\times \frac{1}{{\mathrm{tan}}^{2}\theta}$.

#### Answer:

Given: ……(1)

To Find:

The value of expression

Now, we know that

…… (2)

Now when we compare equation (1) and (2)

We get,

= 512

And

Hypotenuse = 13

Therefore, Triangle representing angle is as shown below

Perpendicular side *AB* is unknown and it can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem

We get,

Therefore by substituting the values of known sides

We get,

Therefore,

Therefore,

…… (3)

Now, we know that

Now from figure (*a*)

We get,

Therefore from figure (*a*) and equation (3) ,

…… (4)

Now we know that,

Therefore, substituting the value of and from equation (1) and (4)

We get,

Therefore 13 gets cancelled and we get

…… (5)

Now we substitute the value of, and from equation (1) , (4) and (5) respectively in the expression below

Therefore,

We get,

Therefore by further simplifying we get,

Now 169 gets cancelled and gets reduced to

Therefore

Therefore the value of is

That is

#### Page No 10.25:

#### Question 31:

If $\mathrm{sec}A=\frac{5}{4}$, verify that $\frac{3\mathrm{sin}A-4{\mathrm{sin}}^{3}A}{4{\mathrm{cos}}^{3}A-3\mathrm{cos}A}=\frac{3\mathrm{tan}A-{\mathrm{tan}}^{3}A}{1-3{\mathrm{tan}}^{2}A}$.

#### Answer:

Given:

…… (1)

To verify:

…… (2)

Now we know that

Therefore

Now, by substituting the value of from equation (1)

We get,

Therefore,

…… (3)

Now, we know the following trigonometric identity

Therefore,

Now by substituting the value of from equation (3)

We get,

Now by taking L.C.M

We get,

Now, by taking square root on both sides

We get,

Therefore,

…… (4)

Now, we know that

Now by substituting the value of and from equation (3) and (4) respectively

We get,

Therefore

…… (5)

Now from the expression of equation (2)

Now by substituting the value of and from equation (3) and (4)

We get,

Therefore,

Now by taking L.C.M of both numerator and denominator

We get,

…… (6)

Now from the expression of equation (2)

Now by substituting the value of from equation (5)

We get,

Now by taking L.C.M

We get,

Now,

Therefore,

Therefore,

…… (7)

Now by comparing equation (6) and (7)

We get,

#### Page No 10.25:

#### Question 32:

If $\mathrm{sin}\theta =\frac{3}{4}$, prove that $\sqrt{\frac{{\mathrm{cosec}}^{2}\theta -{\mathrm{cot}}^{2}\theta}{{\mathrm{sec}}^{2}\theta -1}}=\frac{\sqrt{7}}{3}$.

#### Answer:

Given:

…… (1)

To prove:

…… (2)

By definition,

…… (3)

By Comparing (1) and (3)

We get,

Perpendicular side = 3 and

Hypotenuse = 4

Side *BC* is unknown.

So we find BC by applying Pythagoras theorem to right angled,

Hence,

Now we substitute the value of perpendicular side (*AB*) and hypotenuse (*AC*) and get the base side (*BC*)

Therefore,

Hence, Base side BC = …… (3)

Now,

Therefore from fig. a and equation (3)

Therefore,

…… (4)

Now,

Therefore from fig. a and equation (1) ,

…… (5)

Now,

Therefore from fig. a and equation (4) ,

…… (6)

Now,

Therefore by substituting the values from equation (1) and (4) ,

We get,

Therefore,

…… (7)

Now by substituting the value of , and from equation (5) ,(6) and (7) respectively in the L.H.S of expression (2) ,

We get,

Therefore,

Hence it is proved that

#### Page No 10.25:

#### Question 33:

If $\mathrm{sec}A=\frac{17}{8}$, verify that $\frac{3-4{\mathrm{sin}}^{2}A}{4{\mathrm{cos}}^{2}A-3}=\frac{3-{\mathrm{tan}}^{2}A}{1-3{\mathrm{tan}}^{2}A}$.

#### Answer:

Given:

…… (1)

To verify:

…… (2)

Now we know that

Therefore

Now, by substituting the value of from equation (1)

We get,

Therefore,

…… (3)

Now, we know the following trigonometric identity

Therefore,

Now by substituting the value of from equation (3)

We get,

Now by taking L.C.M

We get,

Now, by taking square root on both sides

We get,

Therefore,

…… (4)

Now, we know that

Now by substituting the value of and from equation (3) and (4) respectively

We get,

Therefore

…… (5)

Now from the expression of equation (2)

Now by substituting the value of and from equation (3) and (4)

We get,

Therefore,

Now by taking L.C.M of both numerator and denominator

We get,

Therefore,

…… (6)

Now from the expression of equation (2)

Now by substituting the value of from equation (5)

We get,

Now by taking L.C.M

We get,

Therefore

Therefore,

…… (7)

Now by comparing equation (6) and (7)

We get,

#### Page No 10.26:

#### Question 34:

If $\mathrm{cot}\theta =\frac{3}{4}$, prove that $\sqrt{\frac{\mathrm{sec}\theta -\mathrm{cosec}\theta}{\mathrm{sec}\theta +\mathrm{cosec}\theta}}=\frac{1}{\sqrt{7}}$.

#### Answer:

Given:

…… (1)

To prove:

Now we know is defined as follows

…… (2)

Now by comparing equation (1) and (2)

We get,

= 3

= 4

Therefore triangle representing angle is as shown below

Side *AC* is unknown and can be found using Pythagoras theorem

Therefore,

Now by substituting the value of known sides from figure

We get,

Now by taking square root on both sides

We get,

Therefore Hypotenuse side *AC *= 5 …… (3)

Now we know, is defined as follows

Therefore from figure (a) and equation (3)

We get,

…… (4)

Now we know

Therefore by substituting the value of from equation (4)

We get,

Therefore,

…… (5)

Now we know, is defined as follows

Therefore from figure (a) and equation (3)

We get,

…… (6)

Now we know

Therefore by substituting the value of from equation (6)

We get,

Therefore,

…… (7)

Now, in expression, by substituting the value of andfrom equation (6) and (7) respectively, we get,

L.C.M of 3 and 4 is 12

Now by taking L.C.M in above expression

We get,

Now 12 gets cancelled and we get,

Now

Therefore,

Now 5 gets cancelled and we get,

Therefore, it is proved that

#### Page No 10.26:

#### Question 35:

If 3 cos θ − 4 sin θ = 2 cos θ + sin θ, find tan θ.

#### Answer:

Given:

To find:

Now consider the given expression

Now by dividing both sides of the above expression by

We get,

Now by separating the denominator for each terms

We get,

Now in the above expression present in both numerator and denominator gets cancelled

Therefore,

…… (1)

Now we know that,

Therefore by substituting in equation (1)

We get,

Now by taking on L.H.S

We get,

Therefore,

Hence

#### Page No 10.26:

#### Question 36:

If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P.

#### Answer:

Given:

To show:

Consider two right angled triangles ABC and PQR such that

Therefore we have,

and

Since it is given that

Therefore,

Now by interchanging position of *AB* and *QR* by cross multiplication

We get,

Let (say) …… (1)

Now by cross multiplication

and …… (2)

Now by using Pythagoras theorem in triangles ABC and PQR

We have,

and

Therefore

and

Now

Now using equation (2)

We get,

Now by taking common

We get,

Therefore,

Now gets cancelled

Therefore,

…… (3)

From (1) and (3)

Therefore,

Hence,

#### Page No 10.41:

#### Question 1:

Evaluate each of the following

sin 45° sin 30° + cos 45° cos 30°

#### Answer:

We have,

…… (1)

Now

So by substituting above values in equation (1)

We get,

Therefore,

#### Page No 10.41:

#### Question 2:

Evaluate each of the following

sin 60 cos 30° + cos 60° sin 30°

#### Answer:

We have to find the value of the expression

…… (1)

Now,

So by substituting above values in equation (1)

We get,

Therefore,

#### Page No 10.41:

#### Question 3:

Evaluate each of the following

cos 60° cos 45° − sin 60° sin 45°

#### Answer:

We have to find the value of the following expression

…… (1)

Now,,

So by substituting above values in equation (1)

We get,

Therefore,

#### Page No 10.41:

#### Question 4:

Evaluate each of the following

sin^{2} 30° + sin^{2} 45° + sin^{2}^{ }60° + sin^{2} 60° + sin^{2} 90°

#### Answer:

We have to find

…… (1)

Now,

,, ,

So by substituting above values in equation (1)

We get,

Now by taking denominator 4 together and simplifying

We get,

Now by taking LCM

We get,

Therefore,

#### Page No 10.41:

#### Question 5:

Evaluate each of the following

cos^{2} 30° + cos^{2} 45° + cos^{2} 60° + cos^{2} 90°

#### Answer:

We have to find the following expression

…… (1)

Now,

,, ,

So by substituting above values in equation (1)

We get,

Now by taking denominator 4 together and simplifying

We get,

Now by taking LCM

We get,

Therefore,

#### Page No 10.41:

#### Question 6:

Evaluate each of the following

tan^{2} 30° + tan^{2} 60° + tan^{2} 45°

#### Answer:

We have to find the following expression

…… (1)

Now,

,,

So by substituting above values in equation (1)

We get,

Now by taking LCM

We get,

Therefore,

#### Page No 10.41:

#### Question 7:

Evaluate each of the following

2 sin^{2} 30° − 3 cos^{2} 45° + tan^{2}^{ }60°

#### Answer:

We have to find the following expression

…… (1)

Now,

,,

So by substituting above values in equation (1)

We get,

In the above equation the first term gets reduced to

Therefore,

In the above equation the first term gets reduced to

Therefore,

Therefore,

#### Page No 10.41:

#### Question 8:

Evaluate each of the following

sin^{2} 30° cos^{2} 45° + 4 tan^{2} 30° + $\frac{1}{2}{\mathrm{sin}}^{2}90\xb0-2{\mathrm{cos}}^{2}90\xb0+\frac{1}{24}{\mathrm{cos}}^{2}0\xb0$

#### Answer:

We have,

…… (1)

Now,

,, ,,

So by substituting above values in equation (1)

We get,

LCM of 8, 3, 2 and 24 is 48

Therefore by taking LCM

We get,

In the above equation the first term gets reduced to

Therefore,

#### Page No 10.42:

#### Question 9:

Evaluate each of the following

4 (sin^{4}^{ }60° + cos^{4} 30°) − 3 (tan^{2} 60° − tan^{2} 45°) + 5 cos^{2} 45°

#### Answer:

We have,

…… (1)

Now,

,, ,

So by substituting above values in equation (1)

We get,

Now, gets reduced to

Therefore,

Now, gets reduced to

Therefore,

Now by taking LCM

We get,

Therefore,

#### Page No 10.42:

#### Question 10:

Evaluate each of the following

(cosec^{2} 45° sec^{2} 30°) (sin^{2} 30° + 4 cot^{2} 45° − sec^{2} 60°)

#### Answer:

We have,

…… (1)

Now,

, , , ,

So by substituting above values in equation (1)

We get,

Now, in above equation 4 cancel 8 and 2 remains

Hence,

Therefore,

#### Page No 10.42:

#### Question 11:

Evaluate each of the following

cosec^{3} 30° cos 60° tan^{3} 45° sin^{2} 90° sec^{2} 45° cot 30°

#### Answer:

We have,

…… (1)

Now,

, , , , ,

So by substituting above values in equation (1)

We get,

Now, 2 gets cancelled and we get,

#### Page No 10.42:

#### Question 12:

Evaluate each of the following

${\mathrm{cot}}^{2}30\xb0-2{\mathrm{cos}}^{2}60\xb0-\frac{3}{4}{\mathrm{sec}}^{2}45\xb0-4{\mathrm{sec}}^{2}30\xb0$

#### Answer:

We have,

…… (1)

Now,

,, ,

So by substituting above values in equation (1)

We get,

Now, in the third term 4 gets cancelled by 2 and 2 remains

Therefore,

Now in the second term, 4 gets cancelled by 2 and 2 remains

Therefore,

Now, LCM of denominator in the above expression is 6

Therefore by taking LCM

We get,

Now in the above expression, gets reduced to

Therefore,

#### Page No 10.42:

#### Question 13:

Evaluate each of the following

(cos 0° + sin 45° + sin 30°) (sin 90° − cos 45° + cos 60°)

#### Answer:

We have,

…… (1)

Now,

,,

So by substituting above values in equation (1)

We get,

Now, LCM of both the product terms in the above expression is

Therefore we get,

Now by rearranging terms in the numerator of above expression

We get,

Now, by applying formula in the numerator of the above expression we get,

…… (2)

Now, we know that

Therefore,

Now, by substituting the above value of in equation (2)

We get,

Now gets reduced to

Therefore,

Hence,

#### Page No 10.42:

#### Question 14:

Evaluate each of the following

$\frac{\mathrm{sin}30\xb0-\mathrm{sin}90\xb0+2\mathrm{cos}0\xb0}{\mathrm{tan}30\xb0\mathrm{tan}60\xb0}$

#### Answer:

We have,

…… (1)

Now,

,,,

So by substituting above values in equation (1)

We get,

Now, present in the denominator of above expression gets cancelled and we get,

Now by taking LCM in the above expression we get,

Therefore,

#### Page No 10.42:

#### Question 15:

Evaluate each of the following

$\frac{4}{{\mathrm{cot}}^{2}30\xb0}+\frac{1}{{\mathrm{sin}}^{2}60\xb0}-{\mathrm{cos}}^{2}45\xb0$

#### Answer:

We have,

…… (1)

Now,

, ,

So by substituting above values in equation (1)

We get,

Now LCM of denominator of above expression is 6

Therefore by taking LCM we get,

Hence,

#### Page No 10.42:

#### Question 16:

Evaluate each of the following

4(sin^{4} 30° + cos^{2} 60°) − 3(cos^{2} 45° − sin^{2} 90°) − sin^{2} 60°

#### Answer:

We have,

…… (1)

Now,

,, ,

So by substituting above values in equation (1)

We get,

LCM of 16 and 4 in the first term of above expression is 16 and

Similarly LCM of 2 and 1 in the second term of above expression is 2

Therefore,

Now in the second term of the above expression

Therefore,

Now, in the above expression 4 cancels 16 and 4 remains in the denominator of first term

Therefore,

Now by taking LCM =4 in the above expression

We get,

Now, in the above expression gets reduced to 2

Therefore,

#### Page No 10.42:

#### Question 17:

Evaluate each of the following

$\frac{{\mathrm{tan}}^{2}60\xb0+4{\mathrm{cos}}^{2}45\xb0+3{\mathrm{sec}}^{2}30\xb0+5{\mathrm{cos}}^{2}90\xb0}{\mathrm{cosec}30\xb0+\mathrm{sec}60\xb0-{\mathrm{cot}}^{2}30\xb0}$

#### Answer:

We have,

…… (1)

Now,

, , , , ,

So by substituting above values in equation (1)

We get,

Now,

3 gets cancel in numerator and we get,

Now, in the numerator get reduced to 2and we get,

Therefore,

#### Page No 10.42:

#### Question 18:

Evaluate each of the following

$\frac{\mathrm{sin}30\xb0}{\mathrm{sin}45\xb0}+\frac{\mathrm{tan}45\xb0}{\mathrm{sec}60\xb0}-\frac{\mathrm{sin}60\xb0}{\mathrm{cot}45\xb0}-\frac{\mathrm{cos}30\xb0}{\mathrm{sin}90\xb0}$

#### Answer:

We have,

…… (1)

Now,

,, ,, ,

So by substituting above values in equation (1)

We get,

Now by further simplifying

We get,

Since,

Therefore,

Now, one gets cancelled and

We get,

Now, by taking LCM

We get,

Therefore,

#### Page No 10.42:

#### Question 19:

Evaluate each of the following

$\frac{\mathrm{tan}45\xb0}{\mathrm{cosec}30\xb0}+\frac{\mathrm{sec}60\xb0}{\mathrm{cot}45\xb0}-\frac{5\mathrm{sin}90\xb0}{2\mathrm{cos}0\xb0}$

#### Answer:

We have,

…… (1)

Now,

,,

So by substituting above values in equation (1)

We get,

Now by taking terms with denominator 2 together and solving

We get,

Now gets reduced to -2

Therefore,

Therefore,

#### Page No 10.42:

#### Question 20:

Find the value of *x* in each of the following :

$2\mathrm{sin}3x=\sqrt{3}$

#### Answer:

We have,

Since,

Therefore,

Therefore,

#### Page No 10.42:

#### Question 21:

Find the value of *x* in each of the following :

$2\mathrm{sin}\frac{x}{2}=1$

#### Answer:

We have,

Since,

Therefore,

Therefore,

#### Page No 10.42:

#### Question 22:

Find the value of *x* in each of the following :

$\sqrt{3}\mathrm{sin}x=\mathrm{cos}x$

#### Answer:

We have,

Now by cross multiplying we get,

…… (1)

Now we know that

…… (2)

Therefore from equation (1) and (2)

We get,

…… (3)

Since,

…… (4)

Therefore, by comparing equation (3) and (4) we get,

Therefore,

#### Page No 10.42:

#### Question 23:

Find the value of *x* in each of the following :

tan *x* = sin 45° cos 45° + sin 30°

#### Answer:

We have,

…… (1)

Now we know that

and

Now by substituting above values in equation (1), we get,

Therefore,

…… (2)

Since,

…… (3)

Therefore by comparing equation (2) and (3)

We get,

Therefore,

#### Page No 10.42:

#### Question 24:

Find the value of *x* in each of the following :

$\sqrt{3}\mathrm{tan}2x=\mathrm{cos}60\xb0+\mathrm{sin}45\xb0\mathrm{cos}45\xb0$

#### Answer:

We have,

…… (1)

Now we know that

and

Now by substituting above values in equation (1), we get,

Therefore,

…… (2)

Since,

…… (3)

Therefore by comparing equation (2) and (3)

We get,

Therefore,

#### Page No 10.42:

#### Question 25:

Find the value of *x* in each of the following :

cos 2*x* = cos 60° cos 30° + sin 60° sin 30°

#### Answer:

We have,

…… (1)

Now we know that

and

Now by substituting above values in equation (1), we get,

Therefore,

Now gets reduced to

Therefore,

…… (2)

Since,

…… (3)

Therefore by comparing equation (2) and (3)

We get,

Therefore,

#### Page No 10.42:

#### Question 26:

If θ = 30°, verify that

(i) $\mathrm{tan}2\mathrm{\theta}=\frac{2\mathrm{tan}\mathrm{\theta}}{1-{\mathrm{tan}}^{2}\mathrm{\theta}}$

(ii) $\mathrm{sin}2\mathrm{\theta}=\frac{2\mathrm{tan}\mathrm{\theta}}{1+{\mathrm{tan}}^{2}\mathrm{\theta}}$

(iii) $\mathrm{cos}2\mathrm{\theta}=\frac{1-{\mathrm{tan}}^{2}\mathrm{\theta}}{1+{\mathrm{tan}}^{2}\mathrm{\theta}}$

(iv) cos 3θ = 4 cos^{3} θ − 3 cos θ

#### Answer:

(i) Given:

…… (1)

To verify:

…… (2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

Now by substituting the value of from equation (1) in the above expression

We get,

Now by substituting the value of from equation (1) in the expression

We get,

…… (4)

Now by comparing equation (3) and (4)

We get,

Hence

(ii) Given:

…… (1)

To verify:

…… (2)

Now consider right hand side

Hence it is verified that,

(iii) Given:

…… (1)

To verify:

…… (2)

Now consider left hand side of the equation (2)

Therefore,

Now consider right hand side of equation (2)

Therefore,

Hence it is verified that,

(iv) Given:

…… (1)

To verify:

…… (2)

Now consider left hand side of the expression in equation (2)

Therefore

Now consider right hand side of the expression to be verified in equation (2)

Therefore,

Hence it is verified that,

#### Page No 10.42:

#### Question 27:

If A = B = 60°, verify that

(i) cos (A − B) = cos A cos B + sin A sin B

(ii) sin (A − B) = sin A cos B − cos A sin B

(iii) $\mathrm{tan}\left(A-B\right)=\frac{\mathrm{tan}A-\mathrm{tan}B}{1+\mathrm{tan}A\mathrm{tan}B}$

#### Answer:

(i) Given:

…… (1)

To verify:

…… (2)

Now consider left hand side of the expression to be verified in equation (2)

Therefore,

Now consider right hand side of the expression to be verified in equation (2)

Therefore,

Hence it is verified that,

(ii) Given:

…… (1)

To verify:

…… (2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

Now by substituting the value of *A* and *B* from equation (1) in the above expression

We get,

Hence it is verified that,

(iii) Given:

…… (1)

To verify:

…… (2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

Now consider RHS of the expression to be verified in equation (2)

Therefore,

Now by substituting the value of *A* and *B* from equation (1) in the above expression

We get,

Hence it is verified that,

#### Page No 10.42:

#### Question 28:

If *A* = 30° and *B* = 60°, verify that

(i) sin (*A* + *B*) = sin *A* cos *B* + cos* A* sin *B*

(ii) cos (*A* + *B*) = cos *A* cos *B* − sin *A* sin *B*

#### Answer:

(i) Given

and …… (1)

To verify:

…… (2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

Now consider RHS of the expression to be verified in equation (2)

Therefore;

$\begin{array}{rcl}\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B& =& \mathrm{sin}30\xb0\mathrm{cos}60\xb0+\mathrm{cos}30\xb0\mathrm{sin}60\xb0\\ & =& \frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\\ & =& \frac{1+3}{4}\\ & =& 1\end{array}$

Hence it is verified that,

(ii) Given:

and …… (1)

To verify:

…… (2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

Now consider RHS of the expression to be verified in equation (2)

Therefore,

Hence it is verified that,

#### Page No 10.43:

#### Question 29:

If sin (A + B) = 1 and cos (A − B) = 1, 0° < A + B ≤ 90°, A ≥ B find A and B.

#### Answer:

Given:

…… (1)

…… (2)

We know that,

…… (3)

…… (4)

Now by comparing equation (1) and (3)

We get,

…… (5)

Now by comparing equation (2) and (4)

We get,

…… (6)

Now to get the values of *A* and *B*, let us solve equation (5) and (6) simultaneously

Therefore by adding equation (5) and (6)

We get,

Therefore,

Hence

Now by subtracting equation (6) from equation (5)

We get,

Therefore,

Hence

Therefore the values of *A* and *B* are as follows

and

#### Page No 10.43:

#### Question 30:

(i) If $\mathrm{tan}\left(A-B\right)=\frac{1}{\sqrt{3}}\mathrm{and}\mathrm{tan}\left(A+B\right)=\sqrt{3},0\xb0A+B\le 90\xb0,AB$ find A and B.

(ii) If $\mathrm{tan}\left(A+B\right)=1\mathrm{and}\mathrm{tan}\left(A-B\right)=\frac{1}{\sqrt{3}},0\xb0A+B90\xb0,AB$, then find the values of *A* and* B*.

#### Answer:

(i)

Given:

…… (1)

…… (2)

We know that,

…… (3)

…… (4)

Now by comparing equation (1) and (3)

We get,

…… (5)

Now by comparing equation (2) and (4)

We get,

…… (6)

Now to get the values of *A* and *B*, let us solve equation (5) and (6) simultaneously

Therefore by adding equation (5) and (6)

We get,

Therefore,

Hence

Now by subtracting equation (5) from equation (6)

We get,

Therefore,

Hence

Therefore the values of *A* and *B* are as follows

and

(ii)

$\mathrm{tan}\left(A+B\right)=1\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\left(A+B\right)=\mathrm{tan}45\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow A+B=45\xb0.....\left(1\right)$

Also,

$\mathrm{tan}\left(A-B\right)=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\left(A-B\right)=\mathrm{tan}30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow A-B=30\xb0.....\left(2\right)$

Adding (1) and (2), we get

$A+B+A-B=45\xb0+30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2A=75\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow A=37.5\xb0$

Putting $A=37.5\xb0$ in (1), we get

$37.5\xb0+B=45\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow B=45\xb0-37.5\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow B=7.5\xb0$

Thus, the values of *A* and *B* are 37.5º and 7.5º, respectively.

#### Page No 10.43:

#### Question 31:

If $\mathrm{sin}\left(A-B\right)=\frac{1}{2}\mathrm{and}\mathrm{cos}\left(A+B\right)=\frac{1}{2},0\xb0A+B\ge 90\xb0,AB$ find A and B.

#### Answer:

Given:

…… (1)

…… (2)

We know that,

…… (3)

…… (4)

Now by comparing equation (1) and (3)

We get,

…… (5)

Now by comparing equation (2) and (4)

We get,

…… (6)

Now to get the values of *A* and *B*, let us solve equation (5) and (6) simultaneously

Therefore by adding equation (5) and (6)

We get,

Therefore,

Hence

Now by subtracting equation (5) from equation (6)

We get,

Therefore,

Hence

Therefore the values of *A* and *B* are as follows

and

#### Page No 10.43:

#### Question 32:

In a ∆ABC right angled at B, ∠A = ∠C. Find the values of

(i) sin A cos C + cos A sin C

(ii) sin A sin B + cos A cos B

#### Answer:

(i) We have drawn the following figure related to given information

To find:

…… (1)

Now we have,

,

,

Now by substituting the above values in equation (1)

We get,

Therefore,

…… (2)

Now in right angled

By applying Pythagoras theorem

We get,

Now, by substituting above value of *AC*^{2} in equation (2)

We get,

Now both numerator and denominator contains

Therefore it gets cancelled and 1 remains

Hence

(ii) We have drawn the following figure

e

To find:

…… (1)

Now we know that sum of all the angles of any triangle is 180°

Therefore,

Since and

Therefore,

It is given that

Therefore,

…… (2)

Now we have,

,

,

Now by substituting the above values in equation (1)

We get,

Since

Therefore

#### Page No 10.43:

#### Question 33:

Find acute angles A and B, if $\mathrm{sin}\left(A+2B\right)=\frac{\sqrt{3}}{2}\mathrm{and}\mathrm{cos}\left(A+4B\right)=0,AB$.

#### Answer:

Given:

…… (1)

…… (2)

We know that,

…… (3)

…… (4)

Now by comparing equation (1) and (3)

We get,

…… (5)

Now by comparing equation (2) and (4)

We get,

…… (6)

Now to get the values of *A* and *B*, let us solve equation (5) and (6) simultaneously

Therefore by subtracting equation (5) from (6)

We get,

Therefore,

Hence

Now by multiplying equation (5) by 2

We get,

…… (7)

Now by subtracting equation (6) from (7)

We get,

Therefore,

Hence

Therefore the values of *A* and *B* are as follows and

#### Page No 10.43:

#### Question 34:

In ∆PQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.

#### Answer:

We are given the following information in the form of triangle

To find: and

Now, in

…… (1)

Now we know that

…… (2)

Now by comparing equation (1) and (2)

We get,

…… (3)

Now we have

Now we know that

Therefore,

Now by cross multiplying

We get,

Therefore,

cm …… (4)

Now we know that

Now we know,

…… (6)

Now by comparing equation (5) and (6)

We get,

…… (7)

Hence from equation (3) and (7)

and

#### Page No 10.43:

#### Question 35:

If sin (*A* − *B*) = sin *A* cos* B* − cos *A *sin *B* and cos (*A* − *B*) = cos *A* cos *B* + sin *A* sin *B*, find the values of sin 15° and cos 15°.

#### Answer:

Given:

…… (1)

…… (2)

To find:

The values of and

In this problem we need to find and

Hence to get angle we need to choose the value of *A* and *B* such that

So If we choose and

Then we get,

Therefore by substituting and in equation (1)

We get,

Therefore,

…… (3)

Now we know that,

, ,

Now by substituting above values in equation (3)

We get,

Therefore,

…… (4)

Now by substituting and in equation (2)

We get,

Therefore,

…… (5)

Now we know that,

, ,

Now by substituting above values in equation (5)

We get,

Therefore,

…… (6)

Therefore from equation (4) and (6)

#### Page No 10.43:

#### Question 36:

In a right triangle ABC, right angled at *C*, if ∠*B* = 60° and *AB* = 15 units. Find the remaining angles and sides.

#### Answer:

We are given the following triangle with related information

It is required to find , and length of sides *AC* and *BC*

is right angled at *C*

Therefore,

Now we know that sum of all the angles of any triangle is

Therefore,

…… (1)

Now by substituting the values of known angles and in equation (1)

We get,

Therefore,

Therefore,

Now,

We know that,

Now we have,

*AB*=15 units and

Therefore by substituting above values in equation (2)

We get,

Now by cross multiplying we get,

Therefore,

…… (3)

Now,

We know that,

Now we have,

AB=15 units and

Therefore by substituting above values in equation (4)

We get,

Now by cross multiplying we get,

Therefore,

Hence,

#### Page No 10.43:

#### Question 37:

If ∆ABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units. Find ∠B, AB and AC.

#### Answer:

We are given the following information in the form of the triangle

It is required to find and length of sides *AB* and *AC*

In

Now we know that sum of all the angles of any triangle is

Therefore,

…… (1)

Now by substituting the values of known angles and in equation (1)

We get,

Therefore,

Therefore,

…… (2)

Now,

We know that,

Now we have,

*BC *= 7 units and

Therefore by substituting above values in equation (3)

We get,

Now by cross multiplying we get,

Therefore,

…… (4)

Now,

We know that,

……(5)

Now we have,

and

Therefore by substituting above values in equation (5)

We get,

Now by cross multiplying we get,

Therefore,

…… (6)

Therefore,

From equation (2), (4) and (6)

, ,

#### Page No 10.43:

#### Question 38:

In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.

#### Answer:

We have drawn the following figure

Since *ABCD* is a rectangle

Therefore,

Now, consider

We know that sum of all the angles of any triangle is

Therefore,

…… (1)

Now by substituting the values of known angles and in equation (1)

We get,

Now in

We know that,

Now we have,

*AB* = 20cm and

Therefore by substituting above values in equation (2)

We get,

Now by cross multiplying we get,

Therefore,

…… (3)

Now in

We know that,

Now we have from equation (3),

*AC*=40cm and

Therefore by substituting above values in equation (4)

We get,

Now by cross multiplying we get,

Therefore,

…… (5)

Since *ABCD* is a rectangle

Therefore,

…… (6)

And

…… (7)

Now in

We know that,

Now by substituting the values of sides from equation (6) and (7)

We get,

Since

Therefore,

That is in

…… (8)

Now in

We know that,

From equation (7)and (8)

Since

Therefore,

Now by cross multiplying we get,

Therefore,

…… (9)

Hence from equation (3), (5) and (9)

#### Page No 10.43:

#### Question 39:

If A and B are acute angles such that $\mathrm{tan}A=\frac{1}{2},\mathrm{tan}B=\frac{1}{3}\mathrm{and}\mathrm{tan}\left(A+B\right)=\frac{\mathrm{tan}A+\mathrm{tan}B}{1-\mathrm{tan}A\mathrm{tan}B}$, find *A* + *B*.

#### Answer:

Given:

…… (1)

…… (2)

…… (3)

Now by substituting the value of and from equation (1) and (2) in equation (3)

We get,

Therefore,

…… (3)

Now we know that

…… (4)

Now by comparing equation (3) and (4)

We get,

#### Page No 10.43:

#### Question 40:

Prove that $\left(\sqrt{3}+1\right)\left(3-cot{30}^{\xb0}\right)={\mathrm{tan}}^{3}{60}^{\xb0}-2\mathrm{sin}{60}^{\xb0}$ .

#### Answer:

Consider the left hand side.

$\left(\sqrt{3}+1\right)\left(3-\mathrm{cot}30\xb0\right)\phantom{\rule{0ex}{0ex}}=\left(\sqrt{3}+1\right)\left(3-\sqrt{3}\right)\left(\mathrm{cot}30\xb0=\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(\sqrt{3}+1\right)\times \sqrt{3}\times \left(\sqrt{3}-1\right)\phantom{\rule{0ex}{0ex}}=\left[{\left(\sqrt{3}\right)}^{2}-1\right]\times \sqrt{3}\phantom{\rule{0ex}{0ex}}=2\sqrt{3}$

Now, consider the right hand side.

${\mathrm{tan}}^{3}60\xb0-2\mathrm{sin}60\xb0\phantom{\rule{0ex}{0ex}}={\left(\sqrt{3}\right)}^{3}-\left(2\times \frac{\sqrt{3}}{2}\right)\left(\mathrm{sin}60\xb0=\frac{\sqrt{3}}{2}\right)\phantom{\rule{0ex}{0ex}}=\sqrt{3}\times \left[{\left(\sqrt{3}\right)}^{2}-1\right]\phantom{\rule{0ex}{0ex}}=2\sqrt{3}$

So, LHS = RHS

Hence proved.

#### Page No 10.52:

#### Question 1:

Evaluate the following :

(i) $\frac{\mathrm{sin}20\xb0}{\mathrm{cos}70\xb0}$

(ii) $\frac{\mathrm{cos}19\xb0}{\mathrm{sin}71\xb0}$

(iii) $\frac{\mathrm{sin}21\xb0}{\mathrm{cos}69\xb0}$

(iv) $\frac{\mathrm{tan}10\xb0}{\mathrm{cot}80\xb0}$

(v) $\frac{\mathrm{sec}11\xb0}{\mathrm{cosec}79\xb0}$

#### Answer:

(i) Given that

Since

Therefore

(ii) Given that

Since

Therefore

(iii) Given that

Since

(iv) We are given that

Since

Therefore

(v) Given that

Since

Therefore

#### Page No 10.53:

#### Question 2:

Evaluate the following :

(i) ${\left(\frac{\mathrm{sin}49\xb0}{\mathrm{cos}41\xb0}\right)}^{2}+{\left(\frac{\mathrm{cos}41\xb0}{\mathrm{sin}49\xb0}\right)}^{2}$

(ii) cos 48° − sin 42°

(iii) $\frac{\mathrm{cot}40\xb0}{\mathrm{tan}50\xb0}-\frac{1}{2}\left(\frac{\mathrm{cos}35\xb0}{\mathrm{sin}55\xb0}\right)$

(iv) ${\left(\frac{\mathrm{sin}27\xb0}{\mathrm{cos}63\xb0}\right)}^{2}-{\left(\frac{\mathrm{cos}63\xb0}{\mathrm{sin}27\xb0}\right)}^{2}$

(v) $\frac{\mathrm{tan}35\xb0}{\mathrm{cot}55\xb0}+\frac{\mathrm{cot}78\xb0}{\mathrm{tan}12\xb0}-1$

(vi) $\frac{\mathrm{sec}70\xb0}{\mathrm{cosec}20\xb0}+\frac{\mathrm{sin}59\xb0}{\mathrm{cos}31\xb0}$

(vii) cosec 31° − sec 59°

(viii) (sin 72° + cos 18°) (sin 72° − cos 18°)

(ix) sin 35° sin 55° − cos 35° cos 55°

(x) tan 48° tan 23° tan 42° tan 67°

(xi) sec50° sin 40° + cos40° cosec 50°

#### Answer:

(i) We have to find:

Since and

So

So value of is

(ii) We have to find:

Since .So

So value of is

(iii) We have to find:

Since and

So value of is

(iv) We have to find:

Since and

So value of is

(v) We have to find:

Since and

So value of is

(vi) We have to find:

Sinceand

So

So value of is

(vii) We have to find:

Since.So

So value of is

(viii) We have to find:

Since.So

So value of is

(ix) We find:

Sinceand

So value of is

(x) We have to find

Since.So

So value of is

(xi) We find to find

Since, and .So

So value of is

#### Page No 10.53:

#### Question 3:

Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°

(i) sin 59° + cos 56°

(ii) tan 65° + cot 49°

(iii) sec 76° + cosec 52°

(iv) cos 78° + sec 78°

(v) cosec 54° + sin 72°

(vi) cot 85° + cos 75°

(vii) sin 67° + cos 75°

#### Answer:

(i) We have and.So

Thus the desired expression is

(ii) We knowand.So

Thus the desired expression is

(iii) We know thatand.So

Thus the desired expression is

(iv) We knowand

Thus the desired expression is

(v) We know and.So

Thus the desired expression is

(vi) We know thatand.So

Thus the desired expression is

(vii) We know thatand.So

Thus the desired expression is

#### Page No 10.53:

#### Question 4:

Express cos 75° + cot 75° in terms of angles between 0° and 30°.

#### Answer:

Given that:

Hence the correct answer is

#### Page No 10.53:

#### Question 5:

If sin 3A = cos (A − 26°), where 3A is an acute angles, find the value of A.

#### Answer:

We are given 3*A* is an acute angle

We have:

Hence the correct answer is

#### Page No 10.53:

#### Question 6:

If A, B, C are the interior angles of a triangle ABC, prove that

(i) $\mathrm{tan}\left(\frac{C+A}{2}\right)=\mathrm{cot}\frac{B}{2}$

(ii) $\mathrm{sin}\left(\frac{B+C}{2}\right)=\mathrm{cos}\frac{A}{2}$

#### Answer:

(i) We have to prove:

Since we know that in triangle

Proved

(ii) We have to prove:

Since we know that in triangle

Proved

#### Page No 10.53:

#### Question 7:

Prove that :

(i) tan 20° tan 35° tan 45° tan 55° tan 70° = 1

(ii) sin 48° sec 42° + cos 48° cosec 42° = 2

(iii) $\frac{\mathrm{sin}70\xb0}{\mathrm{cos}20\xb0}+\frac{\mathrm{cosec}20\xb0}{\mathrm{sec}70\xb0}-2\mathrm{cos}70\xb0\mathrm{cosec}20\xb0=0$

(iv) $\frac{\mathrm{cos}80\xb0}{\mathrm{sin}10\xb0}+\mathrm{cos}59\xb0\mathrm{cosec}31\xb0=2$

#### Answer:

We are asked to find the value of

(i) Therefore

Proved

(ii) We will simplify the left hand side

Proved

(iii) We have,

So we will calculate left hand side

Proved

(iv) We have

We will simplify the left hand side

Proved

#### Page No 10.53:

#### Question 8:

Prove the following :

(i) sin θ sin (90° − θ) − cos θ cos (90° − θ) = 0

(ii) $\frac{\mathrm{cos}\left(90\xb0-\theta \right)\mathrm{sec}\left(90\xb0-\theta \right)\mathrm{tan}\theta}{\mathrm{cosec}\left(90\xb0-\theta \right)\mathrm{sin}\left(90\xb0-\theta \right)\mathrm{cot}\left(90\xb0-\theta \right)}+\frac{\mathrm{tan}\left(90\xb0-\theta \right)}{\mathrm{cot}\theta}=2$

(iii) $\frac{\mathrm{tan}\left(90\xb0-A\right)\mathrm{cot}A}{{\mathrm{cosec}}^{2}A}-{\mathrm{cos}}^{2}A=0$

(iv) $\frac{\mathrm{cos}\left(90\xb0-A\right)\mathrm{sin}\left(90\xb0-A\right)}{\mathrm{tan}\left(90\xb0-A\right)}={\mathrm{sin}}^{2}A$

(v) sin (50° − θ) − cos (40° − θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1

#### Answer:

(i) We have to prove:

Left hand side

=Right hand side

Proved

(ii) We have to prove:

Left hand side

= right hand side

Proved

(iii) We have to prove:

Left hand side

= right hand side

Proved

(iv) We have to prove:

Left hand side

= Right hand side

Proved

(v) We have to prove:

Left hand side

Since .So

=Right hand side

Proved

#### Page No 10.54:

#### Question 9:

Evaluate :

(i) $\frac{2}{3}\left({\mathrm{cos}}^{4}30-{\mathrm{sin}}^{4}45\xb0\right)-3\left({\mathrm{sin}}^{2}60\xb0-{\mathrm{sec}}^{2}45\xb0\right)+\frac{1}{4}{\mathrm{cot}}^{2}30\xb0$

(ii) $4\left({\mathrm{sin}}^{4}30\xb0+{\mathrm{cos}}^{4}60\xb0\right)-\frac{2}{3}\left({\mathrm{sin}}^{2}60\xb0-{\mathrm{cos}}^{2}45\xb0\right)+\frac{1}{2}{\mathrm{tan}}^{2}60\xb0$

(iii) $\frac{\mathrm{sin}50\xb0}{\mathrm{cos}40\xb0}+\frac{\mathrm{cosec}40\xb0}{\mathrm{sec}50\xb0}-4\mathrm{cos}50\xb0\mathrm{cosec}40\xb0$

(iv) tan 35° tan 40° tan 45° tan 50° tan 55°

(v) cosec (65° + θ) − sec (25° − θ) − tan (55° − θ) + cot (35° + θ)

(vi) tan 7° tan 23° tan 60° tan 67° tan 83°

(vii) $\frac{2\mathrm{sin}68\xb0}{\mathrm{cos}22\xb0}-\frac{2\mathrm{cot}15\xb0}{5\mathrm{tan}75\xb0}-\frac{3\mathrm{tan}45\xb0\mathrm{tan}20\xb0\mathrm{tan}40\xb0\mathrm{tan}50\xb0\mathrm{tan}70\xb0}{5}$

(viii) $\frac{3\mathrm{cos}55\xb0}{7\mathrm{sin}35\xb0}-\frac{4\left(\mathrm{cos}70\xb0\mathrm{cosec}20\xb0\right)}{7\left(\mathrm{tan}5\xb0\mathrm{tan}25\xb0\mathrm{tan}45\xb0\mathrm{tan}65\xb0\mathrm{tan}85\xb0\right)}$

(ix) $\frac{\mathrm{sin}18\xb0}{\mathrm{cos}72\xb0}+\sqrt{3}\left\{\mathrm{tan}10\xb0\mathrm{tan}30\xb0\mathrm{tan}40\xb0\mathrm{tan}50\xb0\mathrm{tan}80\xb0\right\}$

(x) $\frac{\mathrm{cos}58\xb0}{\mathrm{sin}32\xb0}+\frac{\mathrm{sin}22\xb0}{\mathrm{cos}68\xb0}-\frac{\mathrm{cos}38\xb0\mathrm{cosec}52\xb0}{\mathrm{tan}18\xb0\mathrm{tan}35\xb0\mathrm{tan}60\xb0\mathrm{tan}72\xb0\mathrm{tan}55\xb0}$

(xi) ${\left(\frac{3\mathrm{tan}41\xb0}{\mathrm{cot}49\xb0}\right)}^{2}-{\left(\frac{\mathrm{sin}35\xb0\mathrm{sec}55\xb0}{\mathrm{tan}10\xb0\mathrm{tan}20\xb0\mathrm{tan}60\xb0\mathrm{tan}70\xb0\mathrm{tan}80\xb0}\right)}^{2}$

#### Answer:

We have to evaluate the following values-

(i) We will use the values of known angles of different trigonometric ratios.

(ii) We will use the values of known angles of different trigonometric ratios.

(iii) We will use the properties of complementary angles.

(iv) We will use the properties of complementary angles.

(v) We will use the properties of complementary angles.

(vi) We will use the properties of complementary angles.

(vii) We will use the properties of complementary angles.

(viii) We will use the properties of complementary angles.

(ix) We will use the properties of complementary angles.

(x) We will use the properties of complementary angles.

(xi)

${\left(\frac{3\mathrm{tan}41\xb0}{\mathrm{cot}49\xb0}\right)}^{2}-{\left(\frac{\mathrm{sin}35\xb0\mathrm{sec}55\xb0}{\mathrm{tan}10\xb0\mathrm{tan}20\xb0\mathrm{tan}60\xb0\mathrm{tan}70\xb0\mathrm{tan}80\xb0}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left[\frac{3\mathrm{tan}\left(90\xb0-49\xb0\right)}{\mathrm{cot}49\xb0}\right]}^{2}-{\left[\frac{\mathrm{sin}35\xb0\mathrm{sec}\left(90\xb0-35\xb0\right)}{\mathrm{tan}10\xb0\mathrm{tan}20\xb0\mathrm{tan}60\xb0\mathrm{tan}\left(90\xb0-20\xb0\right)\mathrm{tan}\left(90\xb0-10\xb0\right)}\right]}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{3\mathrm{cot}49\xb0}{\mathrm{cot}49\xb0}\right)}^{2}-{\left(\frac{\mathrm{sin}35\xb0\mathrm{cosec}35\xb0}{\mathrm{tan}10\xb0\mathrm{tan}20\xb0\times \sqrt{3}\times \mathrm{cot}20\xb0\mathrm{cot}10\xb0}\right)}^{2}\left[\mathrm{tan}\left(90\xb0-\theta \right)=\mathrm{cot}\theta \mathrm{and}\mathrm{sec}\left(90\xb0-\theta \right)=\mathrm{cosec}\theta \right]$

$={\left(3\right)}^{2}-{\left(\frac{1}{\mathrm{tan}10\xb0\mathrm{cot}10\xb0\times \mathrm{tan}20\xb0\mathrm{cot}20\xb0\times \sqrt{3}}\right)}^{2}\left(\mathrm{cosec}\theta =\frac{1}{\mathrm{sin}\theta}\right)\phantom{\rule{0ex}{0ex}}=9-{\left(\frac{1}{1\times 1\times \sqrt{3}}\right)}^{2}\left(\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta}\right)\phantom{\rule{0ex}{0ex}}=9-\frac{1}{3}\phantom{\rule{0ex}{0ex}}=\frac{26}{3}$

#### Page No 10.54:

#### Question 10:

If sin θ = cos (θ − 45°), where θ and θ − 45° are acute angles, find the degree measure of θ.

#### Answer:

Given that: where and are acute angles

We have to find

Therefore

#### Page No 10.54:

#### Question 11:

If A, B, C are the interior angles of a ∆ABC, show that :

(i) $\mathrm{sin}\frac{B+C}{2}=\mathrm{cos}\frac{A}{2}$

(ii) $\mathrm{cos}\frac{B+C}{2}=\mathrm{sin}\frac{A}{2}$

#### Answer:

(i) We have to prove:

Since we know that in triangle

Dividing by 2 on both sides, we get

Proved

(ii) We have to prove:

Since we know that in triangle

Dividing by 2 on both sides, we get

Proved

#### Page No 10.54:

#### Question 12:

If 2θ + 45° and 30° − θ are acute angles, find the degree measure of θ satisfying sin (2θ + 45°) = cos (30° − θ).

#### Answer:

Given that: where and are acute angles

We have to find

So we have

Hence the value of is

#### Page No 10.54:

#### Question 13:

If θ is a positive acute such that sec θ = cosec 60°, find the value of 2 cos^{2} θ − 1.

#### Answer:

We have: where is positive acute angle

Now we have to find

Put

Hence the value of is

#### Page No 10.54:

#### Question 14:

If cos 2θ = sin 4θ, where 2θ and 4θ are acute angles, find the value of θ.

#### Answer:

We have:

Given in question and are acute angles. We have to find

Now we have

#### Page No 10.54:

#### Question 15:

If sin 3 θ = cos (θ − 6°), where 3 θ and θ − 6° are acute angles, find the value of θ.

#### Answer:

We have: where and are acute angles

We have to find

Now we proceed as to find

Therefore* *

#### Page No 10.54:

#### Question 16:

If sec 4*A* = cosec (*A* − 20°), where 4*A* is an acute angles, find the value of* A*.

#### Answer:

Given: and is an acute angle

We have to find

Now

Hence the value of is

#### Page No 10.54:

#### Question 17:

If sec 2*A* = cosec (*A* − 42°), where 2*A* is an acute angles, find the value of* A*.

#### Answer:

Given: and is an acute angle

We have to find

So we proceed as follows to calculate

Hence the value of is

#### Page No 10.54:

#### Question 18:

If tan 2*A* = cot (*A* – 18°), where 2*A* is an acute angle, find the value of *A*.

#### Answer:

Given that,

tan2*A* = cot(*A*− 18°)

cot(90° − 2*A*) = cot(*A* − 18°) (∵ tan*θ* = cot(90° − *θ*))

90° − 2*A* = *A *− 18°

108° = 3*A*

*A*= 36°

#### Page No 10.55:

#### Question 1:

If *θ* is an acute angle such that $\mathrm{cos}\theta =\frac{3}{5},\mathrm{then}\frac{\mathrm{sin}\theta \mathrm{tan}\theta -1}{2{\mathrm{tan}}^{2}\theta}=$

(a) $\frac{16}{625}$

(b) $\frac{1}{36}$

(c) $\frac{3}{160}$

(d) $\frac{160}{3}$

#### Answer:

Given: and we need to find the value of the following expression

We know that:

So we find,

Hence the correct option is

#### Page No 10.55:

#### Question 2:

If $\mathrm{tan}\theta =\frac{a}{b},\mathrm{then}\frac{a\mathrm{sin}\theta +b\mathrm{cos}\theta}{a\mathrm{sin}\theta -b\mathrm{cos}\theta}$is equal to

(a) $\frac{{a}^{2}+{b}^{2}}{{a}^{2}-{b}^{2}}$

(b) $\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}$

(c) $\frac{a+b}{a-b}$

(s) $\frac{a-b}{a+b}$

#### Answer:

Given:

We have to find the value of following expression in terms of *a* and *b*

We know that:

Now we find,

Hence the correct option is

#### Page No 10.55:

#### Question 3:

If 5 tan θ − 4 = 0, then the value of $\frac{5\mathrm{sin}\theta -4\mathrm{cos}\theta}{5\mathrm{sin}\theta +4\mathrm{cos}\theta}$ is

(a) $\frac{5}{3}$

(b) $\frac{5}{6}$

(c) 0

(d) $\frac{1}{6}$

#### Answer:

Given that:.We have to find the value of the following expression

Since

We know that:

Since and

Now we find

Hence the correct option is

#### Page No 10.55:

#### Question 4:

If 16 cot *x* = 12, then $\frac{\mathrm{sin}x-\mathrm{cos}x}{\mathrm{sin}x+\mathrm{cos}x}$ equals

(a) $\frac{1}{7}$

(b) $\frac{3}{7}$

(c) $\frac{2}{7}$

(d) 0

#### Answer:

We are given .We are asked to find the following

We know that:

Now we have

,

We knowand

Now we find

Hence the correct option is

#### Page No 10.55:

#### Question 5:

If 8 tan* x* = 15, then sin *x* − cos *x* is equal to

(a) $\frac{8}{17}$

(b) $\frac{17}{7}$

(c) $\frac{1}{17}$

(d) $\frac{7}{17}$

#### Answer:

Given that:

We know that and

We find:

Hence the correct option is

#### Page No 10.55:

#### Question 6:

If $\mathrm{tan}\theta =\frac{1}{\sqrt{7}},\mathrm{then}\frac{{\mathrm{cosec}}^{2}\theta -{\mathrm{sec}}^{2}\theta}{{\mathrm{cosec}}^{2}\theta +{\mathrm{sec}}^{2}\theta}=$

(a) $\frac{5}{7}$

(b) $\frac{3}{7}$

(c) $\frac{1}{12}$

(d) $\frac{3}{4}$

#### Answer:

Given that:

We are asked to find the value of the following expression

Since

We know that and

We find:

Hence the correct option is

#### Page No 10.56:

#### Question 7:

If $\mathrm{tan}\theta =\frac{3}{4}$, then cos^{2} θ − sin^{2} θ =

(a) $\frac{7}{25}$

(b) 1

(c) $\frac{-7}{25}$

(d) $\frac{4}{25}$

#### Answer:

Given that:

Since

We know that and

We find:

Hence the correct option is

#### Page No 10.56:

#### Question 8:

If θ is an acute angle such that ${\mathrm{tan}}^{2}\theta =\frac{8}{7}$, then the value of $\frac{\left(1+\mathrm{sin}\theta \right)\left(1-\mathrm{sin}\theta \right)}{\left(1+\mathrm{cos}\theta \right)\left(1-\mathrm{cos}\theta \right)}$is

(a) $\frac{7}{8}$

(b) $\frac{8}{7}$

(c) $\frac{7}{4}$

(d) $\frac{64}{49}$

#### Answer:

Given that: andis an acute angle

We have to find the following expression

Since

Since

We know thatand

We find:

Hence the correct option is

#### Page No 10.56:

#### Question 9:

If 3 cos θ = 5 sin θ, then the value of $\frac{5\mathrm{sin}\theta -2{\mathrm{sec}}^{3}\theta +2\mathrm{cos}\theta}{5\mathrm{sin}\theta +2{\mathrm{sec}}^{3}\theta -2\mathrm{cos}\theta}$is

(a) $\frac{271}{979}$

(b) $\frac{316}{2937}$

(c) $\frac{542}{2937}$

(d) None of these

#### Answer:

We have,

So we can manipulate it as,

So now we can get the values of other trigonometric ratios,

So now we will put these values in the equation,

So the answer is (a).

#### Page No 10.56:

#### Question 10:

If tan^{2} 45° − cos^{2} 30° = *x* sin 45° cos 45°, then *x* =

(a) 2

(b) −2

(c) $-\frac{1}{2}$

(d) $\frac{1}{2}$

#### Answer:

We are given:

We have to find *x*

We know that

Hence the correct option is

#### Page No 10.56:

#### Question 11:

The value of cos^{2} 17° − sin^{2} 73° is

(a) 1

(b) $\frac{1}{3}$

(c) 0

(d) −1

#### Answer:

We have:

Hence the correct option is

#### Page No 10.56:

#### Question 12:

The value of $\frac{{\mathrm{cos}}^{3}20\xb0-{\mathrm{cos}}^{3}70\xb0}{{\mathrm{sin}}^{3}70\xb0-{\mathrm{sin}}^{3}20\xb0}$is

(a) $\frac{1}{2}$

(b) $\frac{1}{\sqrt{2}}$

(c) 1

(d) 2

#### Answer:

We have to evaluate the value. The formula to be used,

So,

Now using the properties of complementary angles,

So the answer is

#### Page No 10.56:

#### Question 13:

If $\frac{x{\mathrm{cosec}}^{2}30\xb0{\mathrm{sec}}^{2}45\xb0}{8{\mathrm{cos}}^{2}45\xb0{\mathrm{sin}}^{2}60\xb0}={\mathrm{tan}}^{2}60\xb0-{\mathrm{tan}}^{2}30\xb0$, then *x* =

(a) 1

(b) −1

(c) 2

(d) 0

#### Answer:

We have:

Here we have to find the value of

As we know that

So

Hence the correct option is

#### Page No 10.56:

#### Question 14:

If A and B are complementary angles, then

(a) sin A = sin B

(b) cos A = cos B

(c) tan A = tan B

(d) sec A = cosec B

#### Answer:

Given: and are are complementary angles

Since

Hence the correct option is

#### Page No 10.56:

#### Question 15:

If *x* sin (90° − θ) cot (90° − θ) = cos (90° − θ), then *x* =

(a) 0

(b) 1

(c) −1

(d) 2

#### Answer:

We have:

Here we have to find the value of

We know that

Hence the correct option is

#### Page No 10.56:

#### Question 16:

If *x* tan 45° cos 60° = sin 60° cot 60°, then* x* is equal to

(a) 1

(b) $\sqrt{3}$

(c) $\frac{1}{2}$

(d) $\frac{1}{\sqrt{2}}$

#### Answer:

Given that:

Here we have to find the value of

We know that

Hence the correct option is

#### Page No 10.56:

#### Question 17:

If angles A, B, C to a ∆ABC from an increasing AP, then sin B =

(a) $\frac{1}{2}$

(b) $\frac{\sqrt{3}}{2}$

(c) 1

(d) $\frac{1}{\sqrt{2}}$

#### Answer:

Let the angles of a triangleberespectively which constitute an A.P.As we know that sum of all the three angles of a triangle is. So,

So,

Therefore,

Hence,

So answer is

#### Page No 10.56:

#### Question 18:

If θ is an acute angle such that sec^{2} θ = 3, then the value of $\frac{{\mathrm{tan}}^{2}\theta -{\mathrm{cosec}}^{2}\theta}{{\mathrm{tan}}^{2}\theta +{\mathrm{cosec}}^{2}\theta}$is

(a) $\frac{4}{7}$

(b) $\frac{3}{7}$

(c) $\frac{2}{7}$

(d) $\frac{1}{7}$

#### Answer:

Given that:

We need to find the value of the expression

.So

Here we have to find:

Hence the correct option is

#### Page No 10.57:

#### Question 19:

The value of tan 1° tan 2° tan 3° ...... tan 89° is

(a) 1

(b) −1

(c) 0

(d) None of these

#### Answer:

Here we have to find:

Hence the correct option is

#### Page No 10.57:

#### Question 20:

The value of cos 1° cos 2° cos 3° ..... cos 180° is

(a) 1

(b) 0

(c) −1

(d) None of these

#### Answer:

Here we have to find:

Hence the correct option is

#### Page No 10.57:

#### Question 21:

The value of tan 10° tan 15° tan 75° tan 80° is

(a) −1

(b) 0

(c) 1

(d) None of these

#### Answer:

Here we have to find:

Now

Hence the correct option is

#### Page No 10.57:

#### Question 22:

The value of $\frac{\mathrm{cos}\left(90\xb0-\theta \right)\mathrm{sec}\left(90\xb0-\theta \right)\mathrm{tan}\theta}{\mathrm{cosec}\left(90\xb0-\theta \right)\mathrm{sin}\left(90\xb0-\theta \right)\mathrm{cot}\left(90\xb0-\theta \right)}+\frac{\mathrm{tan}\left(90\xb0-\theta \right)}{\mathrm{cot}\theta}$ is

(a) 1

(b) − 1

(c) 2

(d) −2

#### Answer:

We have to find:

So

Hence the correct option is

#### Page No 10.57:

#### Question 23:

If θ and 2θ − 45° are acute angles such that sin θ = cos (2θ − 45°), then tan θ is equal to

(a) 1

(b) −1

(c) $\sqrt{3}$

(d) $\frac{1}{\sqrt{3}}$

#### Answer:

Given that: and are acute angles

We have to find

Where and are acute angles

Since

Now

Put

Hence the correct option is

#### Page No 10.57:

#### Question 24:

If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ − $\sqrt{3}\mathrm{tan}3\theta $ is equal to

(a) 1

(b) 0

(c) −1

(d) $1+\sqrt{3}$

#### Answer:

We are given that and are acute angles satisfying the following condition

.We are asked to find

Where and are acute angles

Now we have to find:

Hence the correct option is

#### Page No 10.57:

#### Question 25:

If A + B = 90°, then $\frac{\mathrm{tan}A\mathrm{tan}B+\mathrm{tan}A\mathrm{cot}B}{\mathrm{sin}A\mathrm{sec}B}-\frac{{\mathrm{sin}}^{2}B}{{\mathrm{cos}}^{2}A}$ is equal to

(a) cot^{2} A

(b) cot^{2} B

(c) −tan^{2} A

(d) −cot^{2} A

#### Answer:

We have:

We have to find the value of the following expression

So

Hence the correct option is

#### Page No 10.57:

#### Question 26:

$\frac{2\mathrm{tan}30\xb0}{1+{\mathrm{tan}}^{2}30\xb0}$ is equal to

(a) sin 60°

(b) cos 60°

(c) tan 60°

(d) sin 30°

#### Answer:

We have to find the value of the following expression

Hence the correct option is

#### Page No 10.57:

#### Question 27:

$\frac{1-{\mathrm{tan}}^{2}45\xb0}{1+{\mathrm{tan}}^{2}45\xb0}$ is equal to

(a) tan 90°

(b) 1

(c) sin 45°

(d) sin 0°

#### Answer:

We have to find the value of the following

So

We know that

Hence the correct option is

#### Page No 10.57:

#### Question 28:

Sin 2A = 2 sin A is true when A =

(a) 0°

(b) 30°

(c) 45°

(d) 60°

#### Answer:

We are given

So

Hence the correct option is

#### Page No 10.57:

#### Question 29:

$\frac{2\mathrm{tan}30\xb0}{1-{\mathrm{tan}}^{2}30\xb0}$ is equal to

(a) cos 60°

(b) sin 60°

(c) tan 60°

(d) sin 30°

#### Answer:

We are asked to find the value of the following

We know that

Hence the correct option is

#### Page No 10.57:

#### Question 30:

If A, B and C are interior angles of a triangle ABC, then $\mathrm{sin}\left(\frac{B+C}{2}\right)=$

(a) $\mathrm{sin}\frac{A}{2}$

(b) $\mathrm{cos}\frac{A}{2}$

(c) $-\mathrm{sin}\frac{A}{2}$

(d) $-\mathrm{cos}\frac{A}{2}$

#### Answer:

We know that in triangle

So

Hence the correct option is

#### Page No 10.57:

#### Question 31:

If $\mathrm{cos}\theta =\frac{2}{3}$, then 2 sec^{2} θ + 2 tan^{2} θ − 7 is equal to

(a) 1

(b) 0

(c) 3

(d) 4

#### Answer:

Given that:

We have to find

As we are given

We know that:

Now we have to find: .So

Hence the correct option is

#### Page No 10.58:

#### Question 32:

tan 5° ✕ tan 30° ✕ 4 tan 85° is equal to

(a) $\frac{4}{\sqrt{3}}$

(b) $4\sqrt{3}$

(c) 1

(d) 4

#### Answer:

We have to find

We know that

So

Hence the correct option is

#### Page No 10.58:

#### Question 33:

The value of $\frac{\mathrm{tan}55\xb0}{\mathrm{cot}35\xb0}$+ cot 1° cot 2° cot 3° .... cot 90°, is

(a) −2

(b) 2

(c) 1

(d) 0

#### Answer:

We have to find the value of the following expression

Hence the correct option is

#### Page No 10.58:

#### Question 34:

In Fig. 5.47, the value of cos ϕ is

(a) $\frac{5}{4}$

(b) $\frac{5}{3}$

(c) $\frac{3}{5}$

(d) $\frac{4}{5}$

#### Answer:

We should proceed with the fact that sum of angles on one side of a straight line is.

So from the given figure,

So, …… (1)

Now from the triangle,

Now we will use equation (1) in the above,

Therefore,

So the answer is

#### Page No 10.58:

#### Question 35:

In Fig. 5.48, AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ.

(a) $\frac{12}{5}$

(b) $\frac{5}{12}$

(c) $\frac{13}{12}$

(d) $\frac{12}{13}$

#### Answer:

We have the following given data in the figure,

Now we will use Pythagoras theorem in,

Therefore,

So the answer is

#### Page No 10.58:

#### Question 1:

The value of (sin30° + cos30°) – (sin60° + cos60°) is ________.

#### Answer:

$\left(\mathrm{sin}30\xb0+\mathrm{cos}30\xb0\right)-\left(\mathrm{sin}60\xb0+\mathrm{cos}60\xb0\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)-\left(\frac{\sqrt{3}}{2}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}+1}{2}-\frac{\sqrt{3}+1}{2}\phantom{\rule{0ex}{0ex}}=0$

The value of (sin30° + cos30°) – (sin60° + cos60°) is _____0_____.

#### Page No 10.58:

#### Question 2:

The value of $\frac{\mathrm{tan}30\xb0}{\mathrm{cot}60\xb0}$ is ______.

#### Answer:

$\frac{\mathrm{tan}30\xb0}{\mathrm{cot}60\xb0}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{1}{\sqrt{3}}}}{{\displaystyle \frac{1}{\sqrt{3}}}}\phantom{\rule{0ex}{0ex}}=1$

The value of $\frac{\mathrm{tan}30\xb0}{\mathrm{cot}60\xb0}$ is _____1_____.

#### Page No 10.59:

#### Question 3:

The value of (sin 45° + cos 45°)^{3} is _______.

#### Answer:

${\left(\mathrm{sin}45\xb0+\mathrm{cos}45\xb0\right)}^{3}\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)}^{3}\phantom{\rule{0ex}{0ex}}={\left(\frac{2}{\sqrt{2}}\right)}^{3}$

$={\left(\sqrt{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}$

The value of (sin 45° + cos 45°)^{3} is $\overline{)2\sqrt{2}}$.

#### Page No 10.59:

#### Question 4:

The value of (sin 30° + cos 30°)^{2} – (sin 60° – cos 60°)^{2} is __________.

#### Answer:

${\left(\mathrm{sin}30\xb0+\mathrm{cos}30\xb0\right)}^{2}-{\left(\mathrm{sin}60\xb0-\mathrm{cos}60\xb0\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)}^{2}-{\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{3}+1\right)}^{2}-{\left(\sqrt{3}-1\right)}^{2}}{4}$

$=\frac{\left(3+1+2\sqrt{3}\right)-\left(3+1-2\sqrt{3}\right)}{4}\phantom{\rule{0ex}{0ex}}=\frac{4+2\sqrt{3}-4+2\sqrt{3}}{4}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{3}}{4}\phantom{\rule{0ex}{0ex}}=\sqrt{3}$

The value of (sin 30° + cos 30°)^{2} – (sin 60° – cos 60°)^{2} is $\overline{)\sqrt{3}}$.

#### Page No 10.59:

#### Question 5:

The value of (cos^{2} 23° – sin^{2} 67°) is ____________.

#### Answer:

We know that,

$\mathrm{cos}23\xb0=\mathrm{cos}\left(90\xb0-67\xb0\right)=\mathrm{sin}67\xb0\left[\mathrm{cos}\left(90\xb0-\theta \right)=\mathrm{sin}\theta \right]$

$\therefore {\mathrm{cos}}^{2}23\xb0-{\mathrm{sin}}^{2}67\xb0\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{2}67\xb0-{\mathrm{sin}}^{2}67\xb0\phantom{\rule{0ex}{0ex}}=0$

The value of (cos^{2} 23° – sin^{2} 67°) is ________0________.

#### Page No 10.59:

#### Question 6:

The value of the expression $\left\{\frac{{\mathrm{sin}}^{2}22\xb0+{\mathrm{sin}}^{2}68\xb0}{{\mathrm{cos}}^{2}22\xb0+{\mathrm{cos}}^{2}68\xb0}+{\mathrm{sin}}^{2}63\xb0+\mathrm{cos}63\xb0\mathrm{sin}27\xb0\right\}$ is ________.

#### Answer:

We know that,

$\mathrm{sin}68\xb0=\mathrm{sin}\left(90\xb0-22\xb0\right)=\mathrm{cos}22\xb0\left[\mathrm{sin}\left(90\xb0-\theta \right)=\mathrm{cos}\theta \right]\phantom{\rule{0ex}{0ex}}\mathrm{sin}27\xb0=\mathrm{sin}\left(90\xb0-63\xb0\right)=\mathrm{cos}63\xb0\phantom{\rule{0ex}{0ex}}\mathrm{cos}68\xb0=\mathrm{cos}\left(90\xb0-22\xb0\right)=\mathrm{sin}22\xb0\left[\mathrm{cos}\left(90\xb0-\theta \right)=\mathrm{sin}\theta \right]$

$\therefore \frac{{\mathrm{sin}}^{2}22\xb0+{\mathrm{sin}}^{2}68\xb0}{{\mathrm{cos}}^{2}22\xb0+{\mathrm{cos}}^{2}68\xb0}+{\mathrm{sin}}^{2}63\xb0+\mathrm{cos}63\xb0\mathrm{sin}27\xb0\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{2}22\xb0+{\mathrm{cos}}^{2}22\xb0}{{\mathrm{cos}}^{2}22\xb0+{\mathrm{sin}}^{2}22\xb0}+{\mathrm{sin}}^{2}63\xb0+\mathrm{cos}63\xb0\times \mathrm{cos}63\xb0\phantom{\rule{0ex}{0ex}}=\frac{1}{1}+{\mathrm{sin}}^{2}63\xb0+{\mathrm{cos}}^{2}63\xb0\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\right)$

$=1+1\phantom{\rule{0ex}{0ex}}=2$

The value of the expression $\left\{\frac{{\mathrm{sin}}^{2}22\xb0+{\mathrm{sin}}^{2}68\xb0}{{\mathrm{cos}}^{2}22\xb0+{\mathrm{cos}}^{2}68\xb0}+{\mathrm{sin}}^{2}63\xb0+\mathrm{cos}63\xb0\mathrm{sin}27\xb0\right\}$ is ______2______.

#### Page No 10.59:

#### Question 7:

Given that sin α = $\frac{1}{2}$ and cos β = $\frac{1}{2}$, then α + β = _________.

#### Answer:

$\mathrm{sin}\alpha =\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}\alpha =\mathrm{sin}30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha =30\xb0$

Also,

$\mathrm{cos}\beta =\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\beta =\mathrm{cos}60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \beta =60\xb0$

$\therefore \alpha +\beta =30\xb0+60\xb0=90\xb0$

Given that sin α = $\frac{1}{2}$ and cos β = $\frac{1}{2}$, then α + β = ______90°______.

#### Page No 10.59:

#### Question 8:

Given that sin (α – β ) = $\frac{1}{2}$ and cos (α + β) = $\frac{1}{2}$, then α = _______ β = _________.

#### Answer:

$\mathrm{sin}\left(\alpha -\beta \right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}\left(\alpha -\beta \right)=\mathrm{sin}30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha -\beta =30\xb0.....\left(1\right)$

Also,

$\mathrm{cos}\left(\alpha +\beta \right)=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha +\beta =60\xb0.....\left(2\right)$

Adding (1) and (2), we get

$\alpha -\beta +\alpha +\beta =30\xb0+60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\alpha =90\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \alpha =45\xb0$

Putting $\alpha =45\xb0$ in (2), we get

$45\xb0+\beta =60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \beta =60\xb0-45\xb0=15\xb0$

$\therefore \alpha =3\beta $

Given that sin (α – β ) = $\frac{1}{2}$ and cos (α + β) = $\frac{1}{2}$, then α = _____3_____ β = ______45º_____.

#### Page No 10.59:

#### Question 9:

If 4 tan* θ* = 3, then $\frac{4\mathrm{sin}\theta -\mathrm{cos}\theta}{4\mathrm{sin}\theta +\mathrm{cos}\theta}$ is equal to _________.

#### Answer:

$\frac{4\mathrm{sin}\theta -\mathrm{cos}\theta}{4\mathrm{sin}\theta +\mathrm{cos}\theta}$

$=\frac{{\displaystyle \frac{4\mathrm{sin}\theta}{\mathrm{cos}\theta}}-1}{{\displaystyle \frac{4\mathrm{sin}\theta}{\mathrm{cos}\theta}}+1}$ (Dividing numerator and denominator by cos*θ*)

$=\frac{4\mathrm{tan}\theta -1}{4\mathrm{tan}\theta +1}$

$=\frac{3-1}{3+1}\left(\because 4\mathrm{tan}\theta =3\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

If 4 tan* θ* = 3, then $\frac{4\mathrm{sin}\theta -\mathrm{cos}\theta}{4\mathrm{sin}\theta +\mathrm{cos}\theta}$ is equal to $\overline{)\frac{1}{2}}$.

#### Page No 10.59:

#### Question 10:

If cos 5θ = sinθ and 5θ < 90°, then the value of tan 3θ is ___________.

#### Answer:

$\mathrm{cos}5\theta =\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}5\theta =\mathrm{cos}\left(90\xb0-\theta \right)\phantom{\rule{0ex}{0ex}}\Rightarrow 5\theta =90\xb0-\theta \phantom{\rule{0ex}{0ex}}\Rightarrow 6\theta =90\xb0$

$\Rightarrow 3\theta =\frac{90\xb0}{2}=45\xb0\phantom{\rule{0ex}{0ex}}\therefore \mathrm{tan}3\theta =\mathrm{tan}45\xb0=1$

If cos 5θ = sinθ and 5θ < 90°, then the value of tan 3θ is ______1______.

#### Page No 10.59:

#### Question 11:

The value of sec^{2} 60° – tan^{2} 60° is __________.

#### Answer:

${\mathrm{sec}}^{2}60\xb0-{\mathrm{tan}}^{2}60\xb0\phantom{\rule{0ex}{0ex}}={\left(2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}\left(\mathrm{sec}60\xb0=2\mathrm{and}\mathrm{tan}60\xb0=\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=4-3\phantom{\rule{0ex}{0ex}}=1$

The value of sec^{2} 60° – tan^{2} 60° is _______1_______.

#### Page No 10.59:

#### Question 12:

If *A + B* = 90°, then the value of tan^{2} *A* – cot^{2} *B* is _________.

#### Answer:

$A+B=90\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow A=90\xb0-B\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}A=\mathrm{tan}\left(90\xb0-B\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}A=\mathrm{cot}B\left[\mathrm{tan}\left(90\xb0-\theta \right)=\mathrm{cot}\theta \right]$

$\therefore {\mathrm{tan}}^{2}A-{\mathrm{cot}}^{2}B={\mathrm{cot}}^{2}B-{\mathrm{cot}}^{2}B=0\left(\mathrm{tan}A=\mathrm{cot}B\right)$

If *A + B* = 90°, then the value of tan^{2} *A* – cot^{2} *B* is ______0______.

#### Page No 10.59:

#### Question 13:

The value of cos 1° cos 2° cos 3° ...... cos 120° is _______.

#### Answer:

$\mathrm{cos}1\xb0\mathrm{cos}2\xb0\mathrm{cos}3\xb0...\mathrm{cos}120\xb0\phantom{\rule{0ex}{0ex}}=\mathrm{cos}1\xb0\times \mathrm{cos}2\xb0\times \mathrm{cos}3\xb0\times ...\times \mathrm{cos}90\xb0\times ...\times \mathrm{cos}120\xb0\phantom{\rule{0ex}{0ex}}=\mathrm{cos}1\xb0\times \mathrm{cos}2\xb0\times \mathrm{cos}3\xb0\times ...\times 0\times ...\times \mathrm{cos}120\xb0\left(\mathrm{cos}90\xb0=0\right)\phantom{\rule{0ex}{0ex}}=0$

The value of cos 1° cos 2° cos 3° ...... cos 120° is _____0_____.

#### Page No 10.59:

#### Question 14:

If tan θ + cot θ = 2 and, 0° < θ < 90° then tan^{10} θ + cot^{10} θ is equal to _________.

#### Answer:

$\mathrm{tan}\theta +\mathrm{cot}\theta =2\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta +\frac{1}{\mathrm{tan}\theta}=2\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\mathrm{tan}}^{2}\theta +1}{\mathrm{tan}\theta}=2\phantom{\rule{0ex}{0ex}}$

$\Rightarrow {\mathrm{tan}}^{2}\theta +1=2\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{tan}}^{2}\theta -2\mathrm{tan}\theta +1=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\mathrm{tan}\theta -1\right)}^{2}=0$

$\Rightarrow \mathrm{tan}\theta -1=0\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =1\phantom{\rule{0ex}{0ex}}\therefore \mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta}=1$

So,

${\mathrm{tan}}^{10}\theta +{\mathrm{cot}}^{10}\theta ={\left(1\right)}^{10}+{\left(1\right)}^{10}=1+1=2$

If tan θ + cot θ = 2 and, 0° < θ < 90° then tan^{10} θ + cot^{10} θ is equal to ______2______.

#### Page No 10.59:

#### Question 15:

If sin θ + cos θ = 1 and 0° ≤ θ ≤ 90°, then the possible values of θ are ________.

#### Answer:

The given equation is sin*θ* + cos*θ* = 1.

When *θ *= 0°,

LHS = sin*θ* + cos*θ* = sin0° + cos0° = 0 + 1 = 1 = RHS

When *θ *= 90°,

LHS = sin*θ* + cos*θ* = sin90° + cos90° = 1 + 0 = 1 = RHS

Thus, the possible values of *θ *(0° ≤ *θ* ≤ 90°) satisfying the given equation are 0° and 90°.

If sin θ + cos θ = 1 and 0° ≤ θ ≤ 90°, then the possible values of θ are ___0° and 90°___.

#### Page No 10.59:

#### Question 16:

If $\mathrm{sin}\left(A+B\right)=\mathrm{cos}\left(A-B\right)=\frac{\sqrt{3}}{2}$, then cot 2*A* = _________.

#### Answer:

$\mathrm{sin}\left(A+B\right)=\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin}\left(A+B\right)=\mathrm{sin}60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow A+B=60\xb0.....\left(1\right)$

Also,

$\mathrm{cos}\left(A-B\right)=\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}\left(A-B\right)=\mathrm{cos}30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow A-B=30\xb0.....\left(2\right)$

Adding (1) and (2), we get

$A+B+A-B=60\xb0+30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2A=90\xb0\phantom{\rule{0ex}{0ex}}\therefore \mathrm{cot}2A=\mathrm{cot}90\xb0=0$

If $\mathrm{sin}\left(A+B\right)=\mathrm{cos}\left(A-B\right)=\frac{\sqrt{3}}{2}$, then cot 2*A* = ______0______.

#### Page No 10.59:

#### Question 17:

If Δ*ABC* is an isosceles right triangle right angled at *B*, then $\frac{\mathrm{tan}A+\mathrm{cot}C}{\mathrm{cot}A+\mathrm{cot}C}=\_\_\_\_\_\_\_\_\_\_\_\_.$

#### Answer:

It is given that ΔABC is an isosceles right triangle right angled at B.

$\therefore \angle B=90\xb0\mathrm{and}\angle A=\angle C$ .....(1)

In ΔABC,

$A+B+C=180\xb0$ (Angle sum property of a triangle)

$\Rightarrow A+90\xb0+C=180\xb0\left(\mathrm{Given}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow A+C=180\xb0-90\xb0=90\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow A=90\xb0-C.....\left(2\right)$

$\therefore \frac{\mathrm{tan}A+\mathrm{cot}C}{\mathrm{cot}A+\mathrm{cot}C}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{tan}\left(90\xb0-C\right)+\mathrm{cot}C}{\mathrm{cot}A+\mathrm{cot}C}\left[\mathrm{Using}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{cot}C+\mathrm{cot}C}{\mathrm{cot}A+\mathrm{cot}C}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{cot}C}{2\mathrm{cot}C}\left[\mathrm{Using}\left(1\right)\right]$

= 1

If Δ*ABC* is an isosceles right triangle right angled at *B*, then $\frac{\mathrm{tan}A+\mathrm{cot}C}{\mathrm{cot}A+\mathrm{cot}C}=\overline{)1}.$

#### Page No 10.59:

#### Question 18:

If α + β = 90° and $\mathrm{\alpha}=\frac{\mathrm{\beta}}{2}$, then tan α tan β = ________.

#### Answer:

$\alpha +\beta =90\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\beta}{2}+\beta =90\xb0\left[\alpha =\frac{\beta}{2}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3\beta}{2}=90\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \beta =\frac{90\xb0\times 2}{3}=60\xb0$

$\therefore \alpha =\frac{60\xb0}{2}=30\xb0$

Now,

$\mathrm{tan}\alpha \mathrm{tan}\beta =\mathrm{tan}30\xb0\times \mathrm{tan}60\xb0=\frac{1}{\sqrt{3}}\times \sqrt{3}=1$

If α + β = 90° and $\mathrm{\alpha}=\frac{\mathrm{\beta}}{2}$, then tan α tan β = ______1______.

#### Page No 10.59:

#### Question 19:

If in a triangle *ABC*, angles *A* and *B* are complementary, then the value of cot *C* is ________.

#### Answer:

It is given that angles *A* and *B* are complementary.

$\therefore A+B=90\xb0$ .....(1)

In ∆ABC,

$A+B+C=180\xb0$ (Angle sum property of a triangle)

$\Rightarrow 90\xb0+C=180\xb0\left[\mathrm{Using}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow C=180\xb0-90\xb0=90\xb0\phantom{\rule{0ex}{0ex}}\therefore \mathrm{cot}C=\mathrm{cot}90\xb0=0\phantom{\rule{0ex}{0ex}}$

If in a triangle *ABC*, angles *A* and *B* are complementary, then the value of cot *C* is ______0______.

#### Page No 10.59:

#### Question 20:

The value of tan 25° tan 10° tan 80° tan 65° is __________.

#### Answer:

$\mathrm{tan}25\xb0\mathrm{tan}10\xb0\mathrm{tan}80\xb0\mathrm{tan}65\xb0\phantom{\rule{0ex}{0ex}}=\mathrm{tan}25\xb0\mathrm{tan}10\xb0\mathrm{tan}\left(90\xb0-10\xb0\right)\mathrm{tan}\left(90\xb0-25\xb0\right)\phantom{\rule{0ex}{0ex}}=\mathrm{tan}25\xb0\mathrm{tan}10\xb0\mathrm{cot}10\xb0\mathrm{cot}25\xb0\left[\mathrm{tan}\left(90\xb0-\theta \right)=\mathrm{cot}\theta \right]$

$=\mathrm{tan}25\xb0\mathrm{cot}25\xb0\times \mathrm{tan}10\xb0\mathrm{cot}10\xb0\phantom{\rule{0ex}{0ex}}=1\times 1\left(\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta}\right)\phantom{\rule{0ex}{0ex}}=1$

The value of tan 25° tan 10° tan 80° tan 65° is _______1_______.

#### Page No 10.59:

#### Question 1:

Write the maximum and minimum values of sin θ.

#### Answer:

The maximum value of is and the minimum value of is because value of lies between −1 and 1

#### Page No 10.59:

#### Question 2:

Write the maximum and minimum values of cos θ.

#### Answer:

The maximum value of is and the minimum value of is because value of lies between −1 and 1

#### Page No 10.60:

#### Question 3:

What is the maximum value of $\frac{1}{\mathrm{sec}\theta}$?

#### Answer:

The maximum value of is because the maximum value of is that is

#### Page No 10.60:

#### Question 4:

What is the maximum value of $\frac{1}{\mathrm{cosec}\theta}$?

#### Answer:

The maximum value of is because the maximum value of is that is

#### Page No 10.60:

#### Question 5:

If $\mathrm{tan}\theta =\frac{4}{5}$, find the value of $\frac{\mathrm{cos}\theta -\mathrm{sin}\theta}{\mathrm{cos}\theta +\mathrm{sin}\theta}$.

#### Answer:

It is given that .

We have to find .

#### Page No 10.60:

#### Question 6:

If $\mathrm{cos}\theta =\frac{2}{3}$, find the value of $\frac{\mathrm{sec}\theta -1}{\mathrm{sec}\theta +1}$.

#### Answer:

Given in question:

We have to find

Hence the value of is

#### Page No 10.60:

#### Question 7:

If 3 cot θ = 4, find the value of $\frac{4\mathrm{cos}\theta -\mathrm{sin}\theta}{2\mathrm{cos}\theta +\mathrm{sin}\theta}$.

#### Answer:

We have:

Since we know that in right angle triangle

Now, we find

Hence the value of is

#### Page No 10.60:

#### Question 8:

Given $\mathrm{tan}\theta =\frac{1}{\sqrt{5}}$, what is the value of $\frac{{\mathrm{cosec}}^{2}\theta -{\mathrm{sec}}^{2}\theta}{{\mathrm{cosec}}^{2}\theta +{\mathrm{sec}}^{2}\theta}$?

#### Answer:

Given: ,

We know that:

Now we find,

Hence the value of is

#### Page No 10.60:

#### Question 9:

If $\mathrm{cot}\theta =\frac{1}{\sqrt{3}}$, write the value of $\frac{1-{\mathrm{cos}}^{2}\theta}{2-{\mathrm{sin}}^{2}\theta}$.

#### Answer:

Given:

Now we find,

Hence the value of is

#### Page No 10.60:

#### Question 10:

If $\mathrm{tan}A=\frac{3}{4}\mathrm{and}A+B=90\xb0$, then what is the value of cot B?

#### Answer:

Given that:

Hence the value of is

#### Page No 10.60:

#### Question 11:

If A + B = 90° and $\mathrm{cos}B=\frac{3}{5}$, what is the value of sin A?

#### Answer:

We have:

Hence the value of is

#### Page No 10.60:

#### Question 12:

Write the acute angle θ satisfying $\sqrt{3}\mathrm{sin}\theta =\mathrm{cos}\theta $.

#### Answer:

We have:

Hence the acute angle is

#### Page No 10.60:

#### Question 13:

Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.

#### Answer:

Given that:

Hence the value of is

#### Page No 10.60:

#### Question 14:

Write the value of tan 10° tan 15° tan 75° tan 80°?

#### Answer:

We have to find:

Hence the value of is

#### Page No 10.60:

#### Question 15:

If *A* + *B* = 90° and $\mathrm{tan}A=\frac{3}{4}$, what is cot *B*?

#### Answer:

Given in question:

Hence the value of is

#### Page No 10.60:

#### Question 16:

If $\mathrm{tan}A=\frac{5}{12}$, find the value of (sin A + cos A) sec A.

#### Answer:

Given:

We know that:

Now we find,

Hence the value of is

#### Page No 10.60:

#### Question 17:

If Fig. 10.49, *PS *= 3 cm, *QS *= 4 cm, ∠*PRQ *= θ, ∠*PSQ *= 90°, *PQ *⊥ *RQ *and *RQ *= 9 cm. Evaluate tan θ.

#### Answer:

In right ∆PQS,

${\mathrm{PQ}}^{2}={\mathrm{PS}}^{2}+{\mathrm{SQ}}^{2}\left(\mathrm{Pythagoras}\mathrm{theorem}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{PQ}}^{2}={\left(3\right)}^{2}+{\left(4\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{PQ}}^{2}=9+16=25\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{PQ}=\sqrt{25}=5\mathrm{cm}$

In right ∆PQR,

$\mathrm{tan}\theta =\frac{\mathrm{PQ}}{\mathrm{RQ}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\theta =\frac{5\mathrm{cm}}{9\mathrm{cm}}=\frac{5}{9}\phantom{\rule{0ex}{0ex}}$

Thus, the value of tan*θ* is $\frac{5}{9}$.

#### Page No 10.60:

#### Question 18:

Find *A*, if tan 2*A* = cot(*A *– 24°).

#### Answer:

$\mathrm{tan}2A=\mathrm{cot}\left(A-24\xb0\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}2A=\mathrm{tan}\left[90\xb0-\left(A-24\xb0\right)\right]\left[\mathrm{tan}\left(90\xb0-\theta \right)=\mathrm{cot}\theta \right]\phantom{\rule{0ex}{0ex}}\Rightarrow 2A=90\xb0-\left(A-24\xb0\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 2A=90\xb0-A+24\xb0$

$\Rightarrow 3A=114\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow A=\frac{114\xb0}{3}=38\xb0$

Thus, the value of *A* is 38º.

#### Page No 10.60:

#### Question 19:

Find the value of sin^{2}33° + sin^{2}57°.

#### Answer:

We have

$\mathrm{sin}57\xb0=\mathrm{sin}\left(90\xb0-33\xb0\right)=\mathrm{cos}33\xb0\left[\mathrm{sin}\left(90\xb0-\theta \right)=\mathrm{cos}\theta \right]$

$\therefore {\mathrm{sin}}^{2}33\xb0+{\mathrm{sin}}^{2}57\xb0\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{2}33\xb0+{\mathrm{cos}}^{2}33\xb0\phantom{\rule{0ex}{0ex}}=1\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1\right)$

Thus, the value of sin^{2}33° + sin^{2}57° is 1.

#### Page No 10.60:

#### Question 20:

Evaluate sin^{2} 60° + 2tan 45° – cos^{2} 30°.

#### Answer:

${\mathrm{sin}}^{2}60\xb0+2\mathrm{tan}45\xb0-{\mathrm{cos}}^{2}30\xb0\phantom{\rule{0ex}{0ex}}={\left(\frac{\sqrt{3}}{2}\right)}^{2}+2\times 1-{\left(\frac{\sqrt{3}}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{3}{4}+2-\frac{3}{4}\phantom{\rule{0ex}{0ex}}=2$

#### Page No 10.60:

#### Question 21:

If sin $A=\frac{3}{4}$, calculate sec *A*.

#### Answer:

We know

${\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{3}{4}\right)}^{2}+{\mathrm{cos}}^{2}A=1\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{cos}}^{2}A=1-\frac{9}{16}=\frac{7}{16}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{cos}A=\frac{\sqrt{7}}{4}$

Now,

$\mathrm{sec}A=\frac{1}{\mathrm{cos}A}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sec}A=\frac{1}{\left({\displaystyle \frac{\sqrt{7}}{4}}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sec}A=\frac{4}{\sqrt{7}}=\frac{4\sqrt{7}}{7}$

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