NCERT Solutions for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions

Exercise 2.1: In this exercise, students will learn about the inverse trigonometric functions, including their domain, range, and principal values. They will also practice finding the inverse trigonometric functions of various values and solving equations involving inverse trigonometric functions.

Exercise 2.2: This exercise focuses on the properties of inverse trigonometric functions. Students will learn about the properties of sine, cosine, tangent, cotangent, secant, and cosecant functions, and practice using these properties to evaluate inverse trigonometric functions.

Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of inverse trigonometric functions to solve various problems and answer questions. They will also practice finding the values of inverse trigonometric functions and solving equations involving inverse trigonometric functions.

Overall, this chapter is an important topic in calculus and covers the concepts of inverse trigonometric functions, which are useful in solving various types of mathematical problems.

Access NCERT Solutions for Class 12 Maths  Chapter 2 - Inverse Trigonometric Functions

Exercise 2.1

1. The principal value of $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)$ is

Ans: Assuming $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{=x}$

Further solving,

\[\text{sin x=}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\]

$\text{=sin}\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

The principal value of $\text{si}{{\text{n}}^{\text{-1}}}\text{x}$ lies in the range of $\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$

Therefore, the principal value of $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)$ is $\text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

2. The principal value of $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\sqrt{\text{3}}}{\text{2}} \right)$ is

Ans: Assuming $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\sqrt{\text{3}}}{\text{2}} \right)\text{=x}$

Further solving,

$\text{cos x=}\left( \frac{\sqrt{\text{3}}}{\text{2}} \right)$

$\text{=cos}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

The principal value of $\text{co}{{\text{s}}^{\text{-1}}}\text{x}$ lies in the range of $\left[ \text{0, }\!\!\pi\!\!\text{ } \right]$

Therefore, the principal value of $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\sqrt{\text{3}}}{\text{2}} \right)$ is $\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

3. The principal value of $\text{cose}{{\text{c}}^{\text{-1}}}\left( \text{2} \right)$ is

Ans: Assuming $\text{cose}{{\text{c}}^{\text{-1}}}\left( \text{2} \right)\text{=x}$

Further solving,

$\text{cosec(x)=2}$

$\text{=cosec}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

The principal value of $\text{cose}{{\text{c}}^{\text{-1}}}\text{x}$ lies in the range of $\left[ \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right]\text{-}\left\{ \text{0} \right\}$

Therefore, the principal value of $\text{cose}{{\text{c}}^{\text{-1}}}\left( \text{2} \right)$ is $\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

4. The principal value of $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{-}\sqrt{\text{3}} \right)$ is

Ans: Assuming $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{-}\sqrt{\text{3}} \right)\text{=x}$

Further solving,

$\text{tan x=-}\sqrt{\text{3}}$

$\text{=tan}\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

The principal value of $\text{ta}{{\text{n}}^{\text{-1}}}\text{x}$ lies in the range of $\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$

Therefore, the principal value of $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{-}\sqrt{\text{3}} \right)$ is \[\text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

5. The principal value of $\text{co}{{\text{s}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)$ is

Ans: Assuming $\text{co}{{\text{s}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{=x}$

Further solving,

$\text{cos x=-}\frac{\text{1}}{\text{2}}$

$\text{=cos}\left( \frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

The principal value of $\text{co}{{\text{s}}^{\text{-1}}}\text{x}$ lies in the range of $\left[ \text{0, }\!\!\pi\!\!\text{ } \right]$

Therefore, the principal value of $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)$ is $\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$

6. The principal value of $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{-1} \right)$ is

Ans: Assuming $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{-1} \right)\text{=x}$

Further solving,

$\text{tan x=-1}$

$\text{=tan}\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right)$

The principal value of $\text{ta}{{\text{n}}^{\text{-1}}}\text{x}$ lies in the range of $\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$

Therefore, the principal value of $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{-1} \right)$ is $\text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

7. The principal value of $\text{se}{{\text{c}}^{\text{-1}}}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)$ is

Ans: Assuming $\text{se}{{\text{c}}^{\text{-1}}}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)\text{=x}$

Further solving,

$ \text{secx=}\frac{\text{2}}{\sqrt{\text{3}}} $ 

$ \text{=sec}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

The principal value of $\text{se}{{\text{c}}^{\text{-1}}}\text{x}$ lies in the range of $\left[ \text{0, }\!\!\pi\!\!\text{ } \right]$

Therefore, the principal value of $\text{se}{{\text{c}}^{\text{-1}}}\left( \frac{\text{2}}{\sqrt{\text{3}}} \right)$ is $\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

8. The principal value of $\text{co}{{\text{t}}^{\text{-1}}}\left( \sqrt{\text{3}} \right)$ is

Ans: Assuming $\text{co}{{\text{t}}^{\text{-1}}}\left( \sqrt{\text{3}} \right)\text{=x}$

Further solving,

$ \text{cotx=}\sqrt{\text{3}} $ 

$ \text{=cot}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

The principal value of $\text{co}{{\text{t}}^{\text{-1}}}\text{x}$ lies in the range of $\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$

Therefore, the principal value of \[\text{co}{{\text{t}}^{\text{-1}}}\left( \sqrt{\text{3}} \right)\] is $\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

9. The principal value of $\text{co}{{\text{s}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\sqrt{\text{2}}} \right)$ is

Ans: Assuming $\text{co}{{\text{s}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\sqrt{\text{2}}} \right)\text{=x}$

Further solving,

$ \text{cos x=-}\frac{\text{1}}{\sqrt{\text{2}}} $ 

 $ \text{=cos}\left( \frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right) $ 

The principal value of $\text{co}{{\text{s}}^{\text{-1}}}\text{x}$ lies in the range of $\left[ \text{0, }\!\!\pi\!\!\text{ } \right]$

Therefore, the principal value of \[\text{co}{{\text{s}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\sqrt{\text{2}}} \right)\] is $\frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$

10. The principal value of $\text{cose}{{\text{c}}^{\text{-1}}}\left( \text{-}\sqrt{\text{2}} \right)$ is

Ans: Assuming $\text{cose}{{\text{c}}^{\text{-1}}}\left( \text{-}\sqrt{\text{2}} \right)\text{=x}$

Further solving,

$ \text{cosecx=-}\sqrt{\text{2}} $ 

 $ \text{=cosec}\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right) $ 

The principal value of $\text{co}{{\text{s}}^{\text{-1}}}\text{x}$ lies in the range of $\left[ \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right]\text{-}\left\{ \text{0} \right\}$

Therefore, the principal value of \[\text{cose}{{\text{c}}^{\text{-1}}}\left( \text{-}\sqrt{\text{2}} \right)\] is $\text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

11. Evaluate $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{1} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)$

Ans: Assuming that $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{1} \right)\text{=a}$

$ \text{tan a=1} $ 

$  \text{=tan}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{1} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$ -----(1)

Assuming that $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{=b}$

$ \text{cosb=}\frac{\text{1}}{\text{2}} $ 

 $ \text{=cos}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} $ 

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ -----(2)

Assuming that $\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{=c}$

$ \text{sinc=}\frac{\text{1}}{\text{2}} $ 

$ \text{=sin}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $ 

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$ -----(3)

From Equations (1), (2) and (3),

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \text{1} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right) $ 

$ \text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{+}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}\text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $ 

$ \text{=}\frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} $

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{1} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{=}\frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$

12. Evaluate $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{+2si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)$

Ans: Assuming that $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{=a}$

$ \text{cos a=}\frac{\text{1}}{\text{2}} $ 

$ \text{=cos}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} $ 

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ -----(1)

Assuming that $\text{2si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{=b}$

$ \text{sinb=}\frac{\text{1}}{\text{2}} $ 

$ \text{=sin}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $ 

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

$\text{2si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ -----(2)

From Equations (1) and (2),

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{+2si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right) $ 

 $ \text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{+}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} $ 

 $ \text{=}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} $ 

Therefore, \[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{+2si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)\text{=}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}\]

13. Range of $\text{y}$ if  $\text{si}{{\text{n}}^{\text{-1}}}\text{x=y}$ is

(A) $0\le y\le \pi $

(B) $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$

(C) $0\le y \le \pi $

(D) $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$

Ans: Given that \[\text{si}{{\text{n}}^{\text{-1}}}\text{x=y}\]

Since, the range of \[\text{si}{{\text{n}}^{\text{-1}}}\text{x}\] is \[\left[ \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right]\]

Therefore the range of \[\text{y}\] is \[-\frac{\pi }{2}\le y\le \frac{\pi }{2}\]

14. Evaluate $\text{ta}{{\text{n}}^{\text{-1}}}\left( \sqrt{\text{3}} \right)\text{-se}{{\text{c}}^{\text{-1}}}\left( \text{-2} \right)$

Ans: Assuming that $\text{ta}{{\text{n}}^{\text{-1}}}\left( \sqrt{\text{3}} \right)\text{=a}$

$ \text{tan a=}\sqrt{\text{3}} $

 $ \text{=tan}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} $

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \sqrt{\text{3}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$ -----(1)

Assuming that $\text{se}{{\text{c}}^{\text{-1}}}\left( \text{-2} \right)\text{=b}$

$\text{sec b=-2} $

 $ \text{=sec}\left( \frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} \right) $

$\text{se}{{\text{c}}^{\text{-1}}}\left( \text{-2} \right)\text{=}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}}$-----(2)

From Equations (1) and (2),

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \sqrt{\text{3}} \right)\text{-se}{{\text{c}}^{\text{-1}}}\left( \text{-2} \right) $

$ \text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} $

$ \text{=-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} $

Therefore, \[\text{ta}{{\text{n}}^{\text{-1}}}\left( \sqrt{\text{3}} \right)\text{-se}{{\text{c}}^{\text{-1}}}\left( \text{-2} \right)\text{=-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

Exercise 2.2

1. Using inverse trigonometric identities, prove that  $3{{\sin }^{-1}}x={{\sin }^{-1}}\left( 3x-4{{x}^{3}} \right),x\in \left[ -\frac{1}{2},\frac{1}{2} \right]$

Ans: Assuming \[\text{si}{{\text{n}}^{\text{-1}}}\text{x= }\!\!\alpha\!\!\text{ }\]

\[\text{sin }\!\!\alpha\!\!\text{ =x}\]

We know that $\text{sin3 }\!\!\theta\!\!\text{ =3sin }\!\!\theta\!\!\text{ -4si}{{\text{n}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ }$

Consider

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \text{3x-4}{{\text{x}}^{\text{3}}} \right) $

 $ \text{=si}{{\text{n}}^{\text{-1}}}\left( \text{3sin }\!\!\alpha\!\!\text{ -4si}{{\text{n}}^{\text{3}}}\text{ }\!\!\alpha\!\!\text{ } \right) $

 $ \text{=si}{{\text{n}}^{\text{-1}}}\left( \text{sin3 }\!\!\alpha\!\!\text{ } \right) $

 $ \text{=3 }\!\!\alpha\!\!\text{ } $

 $ \text{=3si}{{\text{n}}^{\text{-1}}}\text{x} $

Hence it is proved that \[\text{3si}{{\text{n}}^{\text{-1}}}\text{x=si}{{\text{n}}^{\text{-1}}}\left( \text{3x-4}{{\text{x}}^{\text{3}}} \right)\]

2. Using inverse trigonometric identities, prove that  $3{{\cos }^{-1}}x={{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right),x\in \left[ \frac{1}{2},1 \right]$

Ans: Assuming \[\text{co}{{\text{s}}^{\text{-1}}}\text{x= }\!\!\alpha\!\!\text{ }\]

\[\text{cos }\!\!\alpha\!\!\text{ =x}\]

We know that $\text{cos3 }\!\!\theta\!\!\text{ =4co}{{\text{s}}^{\text{3}}}\text{ }\!\!\theta\!\!\text{ -3cos }\!\!\theta\!\!\text{ }$

Consider

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \text{4}{{\text{x}}^{\text{3}}}\text{-3x} \right) $

$ \text{=co}{{\text{s}}^{\text{-1}}}\left( \text{4co}{{\text{s}}^{\text{3}}}\text{ }\!\!\alpha\!\!\text{ -3cos }\!\!\alpha\!\!\text{ } \right) $

$ \text{=co}{{\text{s}}^{\text{-1}}}\left( \text{cos3 }\!\!\alpha\!\!\text{ } \right) $ 

 $ \text{=3 }\!\!\alpha\!\!\text{ } $

 $ \text{=3co}{{\text{s}}^{\text{-1}}}\text{x} $ 

Hence it is proved that \[\text{3co}{{\text{s}}^{\text{-1}}}\text{x=co}{{\text{s}}^{\text{-1}}}\left( \text{4}{{\text{x}}^{\text{3}}}\text{-3x} \right)\]

3. Using inverse trigonometric identities, prove that ${{\tan }^{-1}}\frac{2}{11}+{{\tan }^{-1}}\frac{7}{24}={{\tan }^{-1}}\frac{1}{2}$

Ans: We know that $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{a} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \text{b} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{a+b}}{\text{1-ab}} \right)$

Consider

$\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{2}}{\text{11}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{7}}{\text{24}} $ 

$\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\frac{\text{2}}{\text{11}}\text{+}\frac{\text{7}}{\text{24}}}{\text{1-}\frac{\text{2}}{\text{11}}\text{.}\frac{\text{7}}{\text{24}}} $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right) $

Hence it is proved that $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{2}}{\text{11}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{7}}{\text{24}}\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{2}} \right)$                    

4. Using inverse trigonometric identities, prove that $2{{\tan }^{-1}}\frac{1}{2}+{{\tan }^{-1}}\frac{1}{7}={{\tan }^{-1}}\frac{31}{17}$

Ans: We know that $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{a} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \text{b} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{a+b}}{\text{1-ab}} \right)$

Also, $\text{2ta}{{\text{n}}^{-1}}x={{\tan }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)$

Consider

$\text{2ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{7}} $

$\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\text{4}}{\text{3}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{7}} $

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{4}}{\text{3}}\text{+}\frac{\text{1}}{\text{7}}}{\text{1-}\left( \frac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{7}} \right)} \right) $

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{31}}{\text{17}} \right) $

Hence it is proved that $\text{2ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{7}}\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\text{31}}{\text{17}}$

5. Simplify ${{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}-1}{x},x\ne 0$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}\text{-1}}{\text{x}}$

Step 1:

Assuming

$\text{x=tan }\!\!\beta\!\!\text{ } $

 $\text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\text{x} $

Step 2:

Considering

 $ \text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}\text{-1}}{\text{x}} \right) $

 $\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\beta\!\!\text{ }}\text{-1}}{\text{tan }\!\!\beta\!\!\text{ }} \right) $

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{sec }\!\!\beta\!\!\text{ -1}}{\text{tan }\!\!\beta\!\!\text{ }} \right) $

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1-cos }\!\!\beta\!\!\text{ }}{\text{sin }\!\!\beta\!\!\text{ }} \right) $

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{ }\!\!\beta\!\!\text{ }}{\text{2}} \right) \right) $

 $ \text{=}\frac{\text{ }\!\!\beta\!\!\text{ }}{\text{2}} $

 $ \text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x} $

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}\text{-1}}{\text{x}}\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}$

6. Simplify ${{\tan }^{-1}}\frac{1}{\sqrt{{{x}^{2}}-1}},\left| x \right|>1$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\sqrt{{{\text{x}}^{\text{2}}}\text{-1}}}$

Step 1:

Assuming

$ \text{x=sec }\!\!\beta\!\!\text{ } $ 

 $ \text{ }\!\!\beta\!\!\text{ =se}{{\text{c}}^{\text{-1}}}\text{x} $ 

Step 2:

$\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\sqrt{{{\text{x}}^{\text{2}}}\text{-1}}} $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\sqrt{\text{se}{{\text{c}}^{\text{2}}}\text{ }\!\!\beta\!\!\text{ -1}}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{tan }\!\!\beta\!\!\text{ }} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{cot }\!\!\beta\!\!\text{ } \right) $ 

 $ \text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{- }\!\!\beta\!\!\text{ } $ 

 $ \text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-se}{{\text{c}}^{\text{-1}}}\text{x} $ 

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\sqrt{{{\text{x}}^{\text{2}}}\text{-1}}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-se}{{\text{c}}^{\text{-1}}}\text{x}$

7. Simplify $\text{ta}{{\text{n}}^{\text{-1}}}\left( \sqrt{\frac{\text{1-cos x}}{\text{1+cos x}}} \right)\text{,x }\!\!\pi\!\!\text{ }$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\frac{\text{1-cos x}}{\text{1+cos x}}}$

$ \text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\frac{\text{1-cos x}}{\text{1+cos x}}} $ 

  $\text{=ta}{{\text{n}}^{\text{-1}}}\sqrt{\frac{\text{2si}{{\text{n}}^{\text{2}}}\left( \frac{\text{x}}{\text{2}} \right)}{\text{2co}{{\text{s}}^{\text{2}}}\left( \frac{\text{x}}{\text{2}} \right)}} $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{x}}{\text{2}} \right) \right) $ 

 $ \text{=}\frac{\text{x}}{\text{2}} $ 

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\frac{\text{1-cos x}}{\text{1+cos x}}}\text{=}\frac{\text{x}}{\text{2}}$

8. Simplify $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{cos x-sin x}}{\text{cos x+sin x}} \right)\text{,0x }\!\!\pi\!\!\text{ }$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{cos x-sin x}}{\text{cos x+sin x}} \right)$

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{cos x-sin x}}{\text{cos x+sin x}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1-tan x}}{\text{1+tan x}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} \right) \right) $ 

 $ \text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x} $ 

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{cos x-sin x}}{\text{cos x+sin x}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-x}$

9. Simplify $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}}\text{,}\left| \text{x} \right|\text{a}$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}}$

Assuming 

$\text{x=asin }\!\!\alpha\!\!\text{ }$ $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}} $ 

 $\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{asin }\!\!\alpha\!\!\text{ }}{\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{a}}^{\text{2}}}\text{si}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }}} \right) $ 

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan }\!\!\alpha\!\!\text{ } \right) $ 

 $ \text{= }\!\!\alpha\!\!\text{ } $ 

 $\text{=si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{a}} \right) $ 

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{\sqrt{{{\text{a}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}}}\text{=si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{a}} \right)$

10. Simplify ${{\tan }^{-1}}\left( \frac{3{{a}^{2}}x-{{x}^{3}}}{{{a}^{3}}-3a{{x}^{2}}} \right),a>0;\frac{-a}{\sqrt{3}}\le x\le \frac{a}{\sqrt{3}}$

Ans: Consider $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}{{\text{a}}^{\text{2}}}\text{x-}{{\text{x}}^{\text{3}}}}{{{\text{a}}^{\text{3}}}\text{-3a}{{\text{x}}^{\text{2}}}} \right)$

Assuming 

$\text{x=atan }\!\!\alpha\!\!\text{ }$

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}{{\text{a}}^{\text{2}}}\text{x-}{{\text{x}}^{\text{3}}}}{{{\text{a}}^{\text{3}}}\text{-3a}{{\text{x}}^{\text{2}}}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}{{\text{a}}^{\text{3}}}\text{tan }\!\!\alpha\!\!\text{ -}{{\text{a}}^{\text{3}}}\text{tan }\!\!\alpha\!\!\text{ }}{{{\text{a}}^{\text{3}}}\text{-3}{{\text{a}}^{\text{3}}}\text{ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan3 }\!\!\alpha\!\!\text{ } \right) $ 

 $\text{=3 }\!\!\alpha\!\!\text{ } $ 

 $ \text{=3ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{a}} \right) $ 

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}{{\text{a}}^{\text{2}}}\text{x-}{{\text{x}}^{\text{3}}}}{{{\text{a}}^{\text{3}}}\text{-3a}{{\text{x}}^{\text{2}}}} \right)\text{=3ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{a}} \right)$

11. Evaluate $\text{ta}{{\text{n}}^{\text{-1}}}\left[ \text{2cos}\left( \text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}} \right) \right]$

Ans: Assuming $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}}\text{=a}$

$ \text{sin a=}\frac{\text{1}}{\text{2}} $ 

$ \text{=sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

$\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$ -----(1)

Further solving,

From equation (1)

$ \text{ta}{{\text{n}}^{\text{-1}}}\left[ \text{2cos}\left( \text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}} \right) \right] $ 

$\text{=ta}{{\text{n}}^{\text{-1}}}\left[ \text{2cos}\left( \text{2 }\!\!\times\!\!\text{ }\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) \right] $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left[ \text{2cos}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right] $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{1} \right) $ 

 $\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left[ \text{2cos}\left( \text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{2}} \right) \right]\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

12. Evaluate $\text{cot}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{a+co}{{\text{t}}^{\text{-1}}}\text{a} \right)$

Ans: Consider 

$ \text{ta}{{\text{n}}^{\text{-1}}}\text{a+co}{{\text{t}}^{\text{-1}}}\text{a} $ 

 $\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} $ 

Further solving,

$ \text{cot}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{a+co}{{\text{t}}^{\text{-1}}}\text{a} \right) $ 

 $\text{=cot}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right) $ 

 $ \text{=0} $ 

Therefore, $\text{cot}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{a+co}{{\text{t}}^{\text{-1}}}\text{a} \right)\text{=0}$

13. Evaluate $\text{tan}\frac{\text{1}}{\text{2}}\left[ \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{2x}}{\text{1-}{{\text{x}}^{\text{2}}}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1+}{{\text{y}}^{\text{2}}}} \right]\text{,}\left| \text{x} \right|\text{1,y0 and xy1}$

Ans: Assuming $\text{x=tan }\!\!\alpha\!\!\text{ }$

Consider 

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2x}}{\text{1-}{{\text{x}}^{\text{2}}}} \right) $ 

 $\text{=si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2tan }\!\!\alpha\!\!\text{ }}{\text{1-ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} \right) $ 

 $ \text{=si}{{\text{n}}^{\text{-1}}}\left( \text{sin2 }\!\!\alpha\!\!\text{ } \right) $ 

 $ \text{=2 }\!\!\alpha\!\!\text{ } $ 

$\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x}$ ----(1)

Consider $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1+}{{\text{y}}^{\text{2}}}} \right)$

Assuming $\text{y=tan }\!\!\beta\!\!\text{ }$

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1+}{{\text{y}}^{\text{2}}}} \right) $ 

$ \text{=co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\beta\!\!\text{ }}{\text{1+ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\beta\!\!\text{ }} \right) $ 

$ \text{=co}{{\text{s}}^{\text{-1}}}\left( \text{cos2 }\!\!\beta\!\!\text{ } \right) $ 

$ \text{=2 }\!\!\beta\!\!\text{ } $ 

$\text{=2ta}{{\text{n}}^{\text{-1}}}\text{y}$ -----(2)

From equations (1) and (2),

$ \text{tan}\frac{\text{1}}{\text{2}}\left[ \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{2x}}{\text{1-}{{\text{x}}^{\text{2}}}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1+}{{\text{y}}^{\text{2}}}} \right] $ 

$ \text{=tan}\frac{\text{1}}{\text{2}}\left[ \text{2ta}{{\text{n}}^{\text{-1}}}\text{x+2ta}{{\text{n}}^{\text{-1}}}\text{y} \right] $ 

 $\text{=tan}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x+ta}{{\text{n}}^{\text{-1}}}\text{y} \right) $ 

 $ \text{=}\frac{\text{x+y}}{\text{1-xy}} $ 

Therefore, $\text{tan}\frac{\text{1}}{\text{2}}\left[ \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{2x}}{\text{1-}{{\text{x}}^{\text{2}}}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{1-}{{\text{y}}^{\text{2}}}}{\text{1+}{{\text{y}}^{\text{2}}}} \right]\text{=}\frac{\text{x+y}}{\text{1-xy}}$

14. Compute the value of $\text{x}$ which satisfies \[\text{sin}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}}\text{+co}{{\text{s}}^{\text{-1}}}\text{x} \right)\text{=1}\]

Ans: We know that $\text{sin}\left( \text{A+B} \right)\text{=sin a cosB+cos a sinB}$

Consider $\text{sin}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}}\text{+co}{{\text{s}}^{\text{-1}}}\text{x} \right)\text{=1}$

$\text{sin}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}} \right)\text{cos}\left( \text{co}{{\text{s}}^{\text{-1}}}\text{x} \right)\text{+cos}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}} \right)\text{sin}\left( \text{co}{{\text{s}}^{\text{-1}}}\text{x} \right)\text{=1}$

$\frac{\text{x}}{\text{5}}\text{+cos}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}} \right)\text{sin}\left( \text{co}{{\text{s}}^{\text{-1}}}\text{x} \right)\text{=1}$ ----(1)

Assuming 

$ \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}}\text{=a} $ 

 $\text{sin a=}\frac{\text{1}}{\text{5}} $ 

 $\text{cos a=}\sqrt{\text{1-}{{\left( \frac{\text{1}}{\text{5}} \right)}^{\text{2}}}} $ 

 $\text{cos a=}\frac{\text{2}\sqrt{\text{6}}}{\text{5}} $ 

 $\text{a=co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{2}\sqrt{\text{6}}}{\text{5}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1}}{\text{5}} \right)\text{=co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{2}\sqrt{\text{6}}}{\text{5}} \right)\] ----(2)

Assuming 

$ \text{co}{{\text{s}}^{\text{-1}}}\text{x=b} $ 

 $ \text{cosb=x} $ 

 $ \text{sinb=}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} $ 

 $ \text{b=si}{{\text{n}}^{\text{-1}}}\left( \sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right) $ 

$\text{co}{{\text{s}}^{\text{-1}}}\text{x=si}{{\text{n}}^{\text{-1}}}\left( \sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right)$ -----(3)

From Equations (1), (2) and (3),

$\frac{\text{x}}{\text{5}}\text{+}\frac{\text{2}\sqrt{\text{6}}}{\text{5}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{=1} $ 

$\text{x+2}\sqrt{\text{6-6}{{\text{x}}^{\text{2}}}}\text{=5} $ 

$ {{\left( \text{5x-1} \right)}^{\text{2}}}\text{=0} $ 

$ \text{x=}\frac{\text{1}}{\text{5}} $ 

Therefore, for \[\text{x=}\frac{\text{1}}{\text{5}}\], \[\text{sin}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}}\text{+co}{{\text{s}}^{\text{-1}}}\text{x} \right)\text{=1}\]

15. Compute the value of $\text{x}$ which satisfies \[\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x-1}}{\text{x-2}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{x+1}}{\text{x+2}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]

Ans: We know that $\text{ta}{{\text{n}}^{\text{-1}}}\text{a+ta}{{\text{n}}^{\text{-1}}}\text{b=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{a+b}}{\text{1-ab}} \right)$

Consider 

$\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x-1}}{\text{x-2}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{x+1}}{\text{x+2}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

 $ \text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{x-1}}{\text{x-2}}\text{+}\frac{\text{x+1}}{\text{x+2}}}{\text{1-}\left( \frac{\text{x-1}}{\text{x-2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{x+1}}{\text{x+2}} \right)} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

 $ \text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2}{{\text{x}}^{\text{2}}}\text{-4}}{\text{-3}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

 $ \frac{\text{2}{{\text{x}}^{\text{2}}}\text{-4}}{\text{-3}}\text{=1} $ 

 $ \text{x= }\!\!\pm\!\!\text{ }\frac{\text{1}}{\sqrt{\text{2}}} $ 

Therefore, for \[\text{x= }\!\!\pm\!\!\text{ }\frac{\text{1}}{\sqrt{\text{2}}}\], $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{x-1}}{\text{x-2}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{x+1}}{\text{x+2}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

16. Evaluate \[\text{si}{{\text{n}}^{\text{-1}}}\left( \text{sin}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\]

Ans: We know that \[\text{sin}\left( \text{ }\!\!\theta\!\!\text{ } \right)\text{=sin}\left( \text{ }\!\!\pi\!\!\text{ - }\!\!\theta\!\!\text{ } \right)\]

Consider \[\text{si}{{\text{n}}^{\text{-1}}}\left( \text{sin}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\]

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \text{sin}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} \right) $ 

 $\text{=si}{{\text{n}}^{\text{-1}}}\left( \text{sin}\left( \text{ }\!\!\pi\!\!\text{ -}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{3}} \right) \right) $ 

$ \text{=si}{{\text{n}}^{\text{-1}}}\left( \text{sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right) \right) $ 

 $\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} $ 

Therefore, \[{{\sin }^{-1}}\left( \sin \frac{2\pi }{3} \right)=\frac{\pi }{3}\]

17. Evaluate \[{{\tan }^{-1}}\left( \tan \left( \frac{3\pi }{4} \right) \right)\]

Ans: We know that \[\text{tan}\left( \text{ }\!\!\theta\!\!\text{ } \right)\text{=-tan}\left( \text{- }\!\!\theta\!\!\text{ } \right)\]

Consider \[\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right)\]

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right) $ 

 $\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{-tan}\left( \text{-}\frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right) $ 

 $\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right) $ 

 $ \text{=-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

Therefore, \[\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}} \right) \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]

18. Evaluate \[\text{tan}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{+co}{{\text{t}}^{\text{-1}}}\frac{\text{3}}{\text{2}} \right)\]

Ans: Assuming $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=a}$

$\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=a} $ 

$ \text{sin a=}\frac{\text{3}}{\text{5}} $ 

 $ \text{cos a=}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

$ \text{cos a=}\frac{\text{4}}{\text{5}} $ 

 $ \text{tan a=}\frac{\text{3}}{\text{4}} $ 

$ \text{a=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right) $ 

$\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)$ ----(1)

$\text{co}{{\text{t}}^{\text{-1}}}\frac{\text{3}}{\text{2}}\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\text{2}}{\text{3}}$ -----(2)

Further solving, 

$ \text{tan}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{+co}{{\text{t}}^{\text{-1}}}\frac{\text{3}}{\text{2}} \right) $ 

 $ \text{=tan}\left( \text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{4}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{2}}{\text{3}} \right) $ 

 $\text{=tan}\left( \text{ta}{{\text{n}}^{\text{-1}}}\frac{\frac{\text{3}}{\text{4}}\text{+}\frac{\text{2}}{\text{3}}}{\text{1-}\left( \frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{2}}{\text{3}} \right)} \right) $ 

$ \text{=tan}\left( \text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{17}}{\text{6}} \right) $ 

 $ \text{=}\frac{\text{17}}{\text{6}} $ 

Therefore, $\text{tan}\left( \text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{+co}{{\text{t}}^{\text{-1}}}\frac{\text{3}}{\text{2}} \right)\text{=}\frac{\text{17}}{\text{6}}$

19. Value of \[\text{cos}\left( \text{co}{{\text{s}}^{\text{-1}}}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\] is

(A) \[\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}\]

(B) \[\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}\]

(C) \[\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

(D) \[\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}\]

Ans: Consider $\text{cos}\left( \text{co}{{\text{s}}^{\text{-1}}}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)$

$ \text{=cos}\left( \text{co}{{\text{s}}^{\text{-1}}}\frac{\text{-7 }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

$ \text{=co}{{\text{s}}^{\text{-1}}}\left[ \text{cos}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}} \right] $ 

 $\text{=}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}} $ 

Therefore, $\text{cos}\left( \text{co}{{\text{s}}^{\text{-1}}}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=}\frac{\text{5 }\!\!\pi\!\!\text{ }}{\text{6}}$

The correct option is B

20. Value of \[\text{sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right) \right)\] is

(A) \[\frac{\text{1}}{\text{2}}\]

(B) \[\frac{\text{1}}{\text{3}}\]

(C) \[\frac{\text{1}}{\text{4}}\]

(D) \[\text{1}\]

Ans: Assuming $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{=a}$

$\text{sin a=-}\frac{\text{1}}{\text{2}} $ 

 $ \text{=-sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

 $\text{=sin}\left( \text{-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

$\text{si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{=-}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

Further solving,

$ \text{sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right) \right) $ 

 $ \text{=sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{+}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

 $ \text{=sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right) $ 

 $ \text{=1} $ 

Therefore, $\text{sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\text{-si}{{\text{n}}^{\text{-1}}}\left( \text{-}\frac{\text{1}}{\text{2}} \right) \right)\text{=1}$

The correct option is D

Miscellaneous Solutions

1. Evaluate \[\text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\frac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\] 

Ans: Consider $\text{cos}\frac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}}$

$ \text{cos}\frac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} $ 

 $ \text{=cos}\left( \text{2 }\!\!\pi\!\!\text{ +}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

 $ \text{=cos}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

Further solving,

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\frac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

 $ \text{=co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) \right) $ 

 $\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $ 

Therefore, $\text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\frac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

2. Evaluate \[\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\] 

Ans: Consider $\text{tan}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

$ \text{tan}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} $

 $ \text{=tan}\left( \text{ }\!\!\pi\!\!\text{ +}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $

 $ \text{=tan}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $

Further solving,

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right) $

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) \right) $

$ \text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\frac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

3. Using inverse trigonometric identities, prove that  $\text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\text{24}}{\text{7}}$

Ans: Assuming \[\text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\alpha\!\!\text{ }\] ----(1)

$ \text{sin}\frac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\frac{\text{3}}{\text{5}} $ 

 $ \text{cos}\frac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

 $ \text{cos}\frac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\frac{\text{4}}{\text{5}} $ 

Hence,

$ \text{sin }\!\!\alpha\!\!\text{ =2}\left( \frac{\text{3}}{\text{5}} \right)\left( \frac{\text{4}}{\text{5}} \right) $ 

 $ \text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{24}}{\text{25}} $ 

 $ \text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{24}}{\text{25}} \right)}^{\text{2}}}} $ 

 $ \text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{7}}{\text{25}} $ 

Therefore, 

\[\text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{24}}{\text{7}}\]

\[\text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{24}}{\text{7}} \right)\]

From Equation (1), it is proved that 

\[\text{2si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{24}}{\text{7}} \right)\]

4. Using inverse trigonometric identities, prove that  $\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{8}}{\text{17}}\text{+si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\text{77}}{\text{36}}$

Ans: Assuming \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{8}}{\text{17}}\text{= }\!\!\alpha\!\!\text{ }\] 

$\text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{8}}{\text{17}} $ 

 $ \text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{8}}{\text{17}} \right)}^{\text{2}}}} $ 

 $ \text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{15}}{\text{17}} $ 

Hence,

Therefore, 

 $ \text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{8}}{\text{15}} $ 

 $ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{15}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{17}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{15}} \right)\] ----(1) 

Assuming that \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }\]

$ \text{sin }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{5}} $ 

 $ \text{cos }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

 $\text{cos }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{5}} $ 

Therefore, 

$ \text{tan }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{4}} $ 

 $ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)\] ----(2) 

Consider 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{17}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\]

From Equations (1) and (2),

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{17}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{8}}{\text{15}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{8}}{\text{15}}\text{+}\frac{\text{3}}{\text{4}}}{\text{1-}\left( \frac{\text{8}}{\text{15}}\text{ }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{4}} \right)} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{77}}{\text{36}} \right) $ 

Hence, it is proved that \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{8}}{\text{17}}\text{+si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\frac{\text{77}}{\text{36}}\]

5. Using inverse trigonometric identities, prove that  $\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{4}}{\text{5}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{=co}{{\text{s}}^{\text{-1}}}\frac{\text{33}}{\text{65}}$

Ans: Assuming \[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{4}}{\text{5}}\text{= }\!\!\alpha\!\!\text{ }\] 

$\text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{4}}{\text{5}} $ 

 $ \text{sin }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{4}}{\text{5}} \right)}^{\text{2}}}} $ 

 $ \text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{3}}{\text{5}} $ 

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{3}}{\text{4}} $ 

 $ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{4}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)\] ----(1) 

Assuming that \[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{= }\!\!\beta\!\!\text{ }\]

$ \text{cos }\!\!\beta\!\!\text{ =}\frac{\text{12}}{\text{13}} $ 

 $ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{12}}{\text{13}} \right)}^{\text{2}}}} $ 

 $ \text{sin }\!\!\beta\!\!\text{ =}\frac{\text{5}}{\text{13}} $ 

Therefore, 

$ \text{tan }\!\!\beta\!\!\text{ =}\frac{\text{5}}{\text{12}} $ 

 $ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\] ----(2) 

Consider 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{4}}{\text{5}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\]

From Equations (1) and (2),

$\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{4}}{\text{5}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{3}}{\text{4}}\text{+}\frac{\text{5}}{\text{12}}}{\text{1-}\left( \frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{12}} \right)} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{56}}{\text{33}} \right) $ 

 $ \text{=co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{33}}{\text{65}} \right) $ 

Hence, it is proved that \[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{4}}{\text{5}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{=co}{{\text{s}}^{\text{-1}}}\frac{\text{33}}{\text{65}}\]

6. Using inverse trigonometric identities, prove that  $\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{+si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{56}}{\text{65}}$

Ans: Assuming \[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{= }\!\!\alpha\!\!\text{ }\] 

$ \text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{12}}{\text{13}} $ 

 $ \text{sin }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{12}}{\text{13}} \right)}^{\text{2}}}} $ 

 $\text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{13}} $ 

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{12}} $ 

 $ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\] ----(1) 

Assuming that \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }\]

$\text{sin }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{5}} $ 

 $ \text{cos }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

 $ \text{cos }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{5}} $ 

Therefore, 

$\text{tan }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{4}} $ 

$ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)\] ----(2) 

Consider 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\]

From Equations (1) and (2),

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{3}}{\text{4}}\text{+}\frac{\text{5}}{\text{12}}}{\text{1-}\left( \frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{12}} \right)} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{56}}{\text{33}} \right) $ 

 $ \text{=si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{56}}{\text{65}} \right) $ 

Hence, it is proved that \[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{+si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{56}}{\text{65}}\]

7. Using inverse trigonometric identities, prove that  $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{63}}{\text{16}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}$

Ans: Assuming \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{= }\!\!\alpha\!\!\text{ }\] 

$ \text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{13}} $ 

 $ \text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{5}}{\text{13}} \right)}^{\text{2}}}} $ 

 $ \text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{12}}{\text{13}} $ 

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{12}} $ 

 $ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\] ----(1) 

Assuming that \[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }\]

$ \text{cos }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{5}} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

 $ \text{sin }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{5}} $ 

Therefore, 

$\text{tan }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{3}} $ 

 $\text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right)\] ----(2) 

Consider 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\]

From Equations (1) and (2),

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{4}}{\text{3}}\text{+}\frac{\text{5}}{\text{12}}}{\text{1-}\left( \frac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{12}} \right)} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{63}}{\text{16}} \right) $ 

Hence, it is proved that \[\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{63}}{\text{16}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\]

8. Evaluate $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{7}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{3}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{8}}$

Ans: We know that \[\text{ta}{{\text{n}}^{\text{-1}}}\text{a+ta}{{\text{n}}^{\text{-1}}}\text{b=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{a+b}}{\text{1-ab}} \right)\]

Consider \[\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{7}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{3}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{8}}\]

$\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{7}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{3}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{8}} $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{1}}{\text{5}}\text{+}\frac{\text{1}}{\text{7}}}{\text{1-}\left( \frac{\text{1}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{7}} \right)} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{1}}{\text{3}}\text{+}\frac{\text{1}}{\text{8}}}{\text{1-}\left( \frac{\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{8}} \right)} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{6}}{\text{17}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{11}}{\text{23}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{325}}{\text{325}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{1} \right) $ 

 $ \text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

Therefore, \[\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{5}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{7}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{3}}\text{+ta}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{8}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]

9. Using inverse trigonometric identities, prove that  ${{\tan }^{-1}}\sqrt{x}=\frac{1}{2}{{\cos }^{-1}}\left( \frac{1-x}{1+x} \right),x\in \left[ 0,1 \right]$

Ans: Assuming \[\text{x=ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }\] 

Consider

$\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}} $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\text{tan }\!\!\alpha\!\!\text{ } $ 

 $\text{= }\!\!\alpha\!\!\text{ } $ 

\[\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{= }\!\!\alpha\!\!\text{ }\]----(1)

Consider 

$\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right) $ 

 $ \text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }}{\text{1+ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} \right) $ 

 $ \text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{cos2 }\!\!\alpha\!\!\text{ } $ 

 $ \text{= }\!\!\alpha\!\!\text{ } $ 

\[\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right)\text{= }\!\!\alpha\!\!\text{ }\] -----(2)

From Equations (1) and (2),

It is proved that \[\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right)\]

10. Using inverse trigonometric identities, prove that  ${{\cot }^{-1}}\left( \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)=\frac{x}{2},x\in \left( 0,\frac{\pi }{4} \right)$

Ans: We know that  

$\sqrt{\text{1+sin x}}\text{=cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} $ 

 $ \sqrt{\text{1-sin x}}\text{=cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} $ 

Consider

$\text{co}{{\text{t}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+sin x}}\text{+}\sqrt{\text{1-sin x}}}{\sqrt{\text{1+sin x}}\text{-}\sqrt{\text{1-sin x}}} \right) $ 

 $ \text{=co}{{\text{t}}^{\text{-1}}}\left( \frac{\left( \text{cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} \right)\text{+}\left( \text{cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} \right)}{\left( \text{cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} \right)\text{-}\left( \text{cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} \right)} \right) $ 

$ \text{co}{{\text{t}}^{\text{-1}}}\left( \text{cot}\frac{\text{x}}{\text{2}} \right) $ 

 $ \text{=}\frac{\text{x}}{\text{2}} $ 

It is proved that \[\text{co}{{\text{t}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+sin x}}\text{+}\sqrt{\text{1-sin x}}}{\sqrt{\text{1+sin x}}\text{-}\sqrt{\text{1-sin x}}} \right)\text{=}\frac{\text{x}}{\text{2}}\]

11. Using inverse trigonometric identities, prove that  ${{\tan }^{-1}}\left( \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right)=\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x,-\frac{1}{\sqrt{2}}\le x\le 1$

Ans: Assuming

$\text{x=cos2 }\!\!\beta\!\!\text{ } $ 

 $ \sqrt{\text{1+x}} $ 

 $ \text{=}\sqrt{\text{1+cos2 }\!\!\beta\!\!\text{ }} $ 

 $ \text{=}\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ } $ 

 $ \sqrt{\text{1-x}} $ 

 $ \text{=}\sqrt{\text{1-cos2 }\!\!\beta\!\!\text{ }} $ 

 $ \text{=}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ } $ 

Consider

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+x}}\text{-}\sqrt{\text{1-x}}}{\sqrt{\text{1+x}}\text{+}\sqrt{\text{1-x}}} \right) $ 

 $\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ -}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ }}{\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ +}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ }} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{1-tan }\!\!\beta\!\!\text{ }}{\text{1+tan }\!\!\beta\!\!\text{ }} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{- }\!\!\beta\!\!\text{ } \right) \right) $ 

 $\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{- }\!\!\beta\!\!\text{ } $ 

$ \text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{x} $ 

It is proved that \[\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+x}}\text{-}\sqrt{\text{1-x}}}{\sqrt{\text{1+x}}\text{+}\sqrt{\text{1-x}}} \right)\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{x}\]

12. Using inverse trigonometric identities, prove that $\frac{\text{9 }\!\!\pi\!\!\text{ }}{\text{8}}\text{-}\frac{\text{9}}{\text{4}}\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{3}}\text{=}\frac{\text{9}}{\text{4}}\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2}\sqrt{\text{2}}}{\text{3}} \right)$

Ans: Consider

$ \frac{\text{9 }\!\!\pi\!\!\text{ }}{\text{8}}\text{-}\frac{\text{9}}{\text{4}}\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{3}} $ 

$ \text{=}\frac{\text{9}}{\text{4}}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{3}} \right) $ 

$\text{=}\frac{\text{9}}{\text{4}}\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{1}}{\text{3}} $ 

Assuming 

$ \text{co}{{\text{s}}^{\text{-1}}}\frac{\text{1}}{\text{3}}\text{= }\!\!\alpha\!\!\text{ } $ 

 $ \text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{1}}{\text{3}} $ 

 $\text{sin }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{1}}{\text{3}} \right)}^{\text{2}}}} $ 

 $ \text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{2}\sqrt{\text{2}}}{\text{3}} $ 

 $ \text{ }\!\!\alpha\!\!\text{ =si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2}\sqrt{\text{2}}}{\text{3}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{1}}{\text{3}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{2}\sqrt{\text{2}}}{\text{3}}\]

Further solving,

$ \frac{\text{9 }\!\!\pi\!\!\text{ }}{\text{8}}\text{-}\frac{\text{9}}{\text{4}}\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{3}} $ 

$\text{=}\frac{\text{9}}{\text{4}}\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{1}}{\text{3}} $ $\text{=}\frac{\text{9}}{\text{4}}\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{2}\sqrt{\text{2}}}{\text{3}} $ 

Therefore, It is proved that \[\frac{\text{9 }\!\!\pi\!\!\text{ }}{\text{8}}\text{-}\frac{\text{9}}{\text{4}}\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{1}}{\text{3}}\text{=}\frac{\text{9}}{\text{4}}\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{2}\sqrt{\text{2}}}{\text{3}}\]

13. For what value of $x$ does the equation $\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)$ satisfy?

Ans: Consider \[\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)\]

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{2cos x}}{\text{1-co}{{\text{s}}^{\text{2}}}\text{x}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right) $ 

 $ \frac{\text{2cos x}}{\text{1-co}{{\text{s}}^{\text{2}}}\text{x}}\text{=2cosecx} $ 

 $ \text{sin xcos x=si}{{\text{n}}^{\text{2}}}\text{x} $ 

 $ \text{sin x}\left( \text{cos x-sin x} \right)\text{=0} $ 

 $ \text{sin x=0,cos x-sin x=0} $ 

 $ \text{x=0,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

Therefore, \[\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)\] is satisfied for \[\text{x=0,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]

14. For what value of $x$ does the equation $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1-x}}{\text{1+x}}\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x,}\left( \text{x > 0} \right)$ satisfy?

Ans: Consider \[\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1-x}}{\text{1+x}}\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}\]

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-ta}{{\text{n}}^{\text{-1}}}\text{x} \right) \right)\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

 $ \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-ta}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

 $ \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=}\frac{\text{3}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

 $ \text{ta}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $ 

 $ \text{x=}\frac{\text{1}}{\sqrt{\text{3}}} $ 

Therefore, \[\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{1-x}}{\text{1+x}}\text{=}\frac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}\] is satisfied for \[\text{x=}\frac{\text{1}}{\sqrt{\text{3}}}\] 

15. The expression $\text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right)\text{,}\left| \text{x} \right| < \text{1}$ is equal to

(A) $\frac{\text{x}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}$

(B) $\frac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}$

(C) $\frac{\text{1}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}$

(D) $\frac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}$

Ans: Assuming 

$ \text{x=tan }\!\!\beta\!\!\text{ } $ 

 $ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

$ \text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right) $ 

 $ \text{=sin }\!\!\beta\!\!\text{ } $ 

 $ \text{=}\frac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}} $ 

Hence, \[\text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right)\text{=}\frac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}\]

Therefore, the correct option is option D

16. For what value of $x$ does the equation $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ satisfy?

(A) $\text{0,}\frac{\text{1}}{\text{2}}$

(B) \[\text{1,}\frac{\text{1}}{\text{2}}\]

(C) \[0\]

(D) \[\frac{\text{1}}{\text{2}}\]

Ans: Consider

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} $ 

 $\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right) $ 

\[\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=co}{{\text{s}}^{\text{-1}}}\left( \text{1-x} \right)\] -----(1)

Assuming 

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \text{1-x} \right)\text{= }\!\!\beta\!\!\text{ } $ 

 $ \text{cos }\!\!\beta\!\!\text{ =1-x} $ 

 $ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \text{1-x} \right)}^{\text{2}}}} $ 

 $ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

 $ \text{ }\!\!\beta\!\!\text{ =si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\text{1-x=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}}\] -----(2)

Substituting Equation (2) in (1)

$\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

 $ \text{si}{{\text{n}}^{\text{-1}}}\left( \text{-2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right)\text{=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

$\text{-2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{=}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

$\text{4}{{\text{x}}^{\text{2}}}\text{-4}{{\text{x}}^{\text{4}}}\text{=2x-}{{\text{x}}^{\text{2}}} $ 

 $ \text{4}{{\text{x}}^{\text{4}}}\text{-5}{{\text{x}}^{\text{2}}}\text{+2x=0} $ 

 $ \text{x=0,}\frac{\text{1}}{\text{2}}\text{,}\frac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{17}}}{\text{4}} $ 

But considering the given options and also when \[\text{x=}\frac{\text{1}}{\text{2}}\], the equation \[\text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\] doesn’t satisfy

Thus, $\text{x=0}$ is the only solution.

Hence the correct option is C

17. Solving $\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{y}} \right)\text{-ta}{{\text{n}}^{\text{-1}}}\frac{\text{x-y}}{\text{x+y}}$ results in

(A)$\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$

(B) $\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}$

(C) $\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

(D) $\text{-}\frac{\text{3 }\!\!\pi\!\!\text{ }}{\text{4}}$

Ans: Consider

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{y}} \right)\text{-ta}{{\text{n}}^{\text{-1}}}\frac{\text{x-y}}{\text{x+y}} $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{x}}{\text{y}}\text{-}\frac{\text{x-y}}{\text{x+y}}}{\text{1+}\left( \frac{\text{x}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\left( \frac{\text{x-y}}{\text{x+y}} \right) \right)} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{{{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}}{{{\text{y}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\text{1} $ 

 $ \text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

Therefore, \[\text{ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{x}}{\text{y}} \right)\text{-ta}{{\text{n}}^{\text{-1}}}\frac{\text{x-y}}{\text{x+y}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]

Hence, the correct option is option C

Inverse Trigonometric Functions Class 12 CBSE NCERT Solutions

NCERT Solutions for Class 12 Maths – Free PDF Download

Are you looking for NCERT Solutions for Class 12 Maths Inverse Trigonometric Functions PDF online? Learners can find the proper answers of Inverse Trigonometric Functions Class 12 on Vedantu which have been developed by accomplished mentors complimenting their dexterity. The Class 12 Maths CH 2 NCERT Solutions in PDF is uploaded on the official website. Download the file and start solving right away.

NCERT Solutions for Class 12 Maths Other Chapters PDF Download

• Chapter 1 - Relations and Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

NCERT Solutions for Class 12 Maths Chapter 2 All Exercises

Chapter 2 – Inverse Trigonometric Functions NCERT Solutions for Class 12 Maths Chapter 2 

Introduction: In NCERT Class 12th Maths Chapter 2, students will delve deeper into the concepts of Inverse Trigonometric Functions Class 12 NCERT Solutions. One will have to assess what they learnt in the previous chapter of relations and functions and the trigonometric functions studied in Class 11. Therefore, the assessment will pave the way for a deeper understanding of the foundational base of Class 12 Maths Chapter 2 NCERT solutions. Students will learn the restrictions on domains and ranges of trigonometric functions that ensure their inverses and observe their behaviour through graphical representations. A student needs to revise the previous sections of trigonometric functions to understand NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions. Keeping in mind that trigonometric functions are not one-one and onto over their natural domains and ranges and their inverses do not exist, you can go forward. 

Other than this, a refined understanding of Chapter 2 can be found by looking at the provided examples. The questions will encourage a learner to think outside the box and gain a better approach to apprehending Inverse Trigonometric Functions. The level of difficulty will gradually accelerate as one keeps on solving. It will help you get a full idea of what Inverse Trigonometric Functions Class 12 Solutions are all about.  

2.2 Basic Concepts

A student has to reflect on what they studied in Class 11 Trigonometric Functions and in Chapter 1 of Relations and Functions to get a better grasp on Chapter 2. Corresponding invertible functions from the previous chapter paves the way for obtaining graphs of functions. The concepts also make it clear that the graph of an inverse function can be extracted from the correlated graph of the original function as a mirror image. This will further pave the way for understanding the domain and the co-domain that adheres to the principles of functions. 

Exercise 2.1 Solutions: 14 Questions (14 Short Questions) 

2.3 Properties of Inverse Trigonometric Functions

This particular section deals with proving some essential properties of inverse trigonometry Class 12 functions. They are the inverse functions of cosine, sine, tangent, cotangent, secant and cosecant functions with restricted domains. The results are valid within the principal value branches of the corresponding Inverse Trigonometric Functions and wherever defined. The properties of Inverse Trigonometric Functions assist in the calculation of an angle. Some results may not be valid for all values of the domains of Inverse Trigonometric Functions. They will be considered justifiable only for some values of x, for which Inverse Trigonometric Functions are defined.

Importance of Inverse Trigonometric Functions

In the fields of engineering, construction, and architecture, inverse trigonometric ratios are widely used. Inverse trigonometric ratios are the easiest way to find an unknown angle, therefore we utilize them in places where we need to know the angle for our help and quickly get the correct values. 

The following are a few examples of inverse trigonometric ratios in use.

• A right-angled triangle's unknown angles are measured using this formula.

• Used to determine the angle of inclination or depression. 

• It is used by architects to measure the angle of a bridge and its supports.

• Carpenters use this to achieve a particular cut angle.

Exercise 2.2 Solutions: 20 Questions (4 Short Questions, 16 Long Questions) 

Key Features of NCERT Solutions for Class 12 Maths Chapter 2 

It would be best if there is a thorough revision of the concepts to attain good marks in the exam. If the learner adheres to the Class 12 Inverse Trigonometric Functions Class 12 NCERT solutions that have been developed by the experts in NCERT, then they will be able to learn the foundational base and the core concepts. The key features are provided as follows:

• The Inverse Trigonometry Class 12 NCERT solutions are written to assist in studying and learning the chapter. 

• With the Inverse Trigo Class 12 answers, students will be able to implant solid principles required to solve the exercises much quicker. 

• Class 12 Maths NCERT Solutions Chapter 2 help students in case of doubts clearing. 

• Students are expected to follow the instructions provided in the solutions to score better marks in the exam.