NCERT Solutions 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.

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

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

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

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

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

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

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

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

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

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

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

13. 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

14. 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. 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 \[\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right)\]

9. 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 \[\text{co}{{\text{t}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+sin x}}\text{+}\sqrt{\text{1-sin x}}}{\sqrt{\text{1+sin x}}\text{-}\sqrt{\text{1-sin x}}} \right)\text{=}\frac{\text{x}}{\text{2}}\]

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

11. 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 \[\text{x=0,}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]

12. 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 \[\text{x=}\frac{\text{1}}{\sqrt{\text{3}}}\]

13. 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

14. 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

Inverse Trigonometric Functions Class 12 CBSE NCERT Solutions

2.1 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 Inverse trigonometric functions class 12 NCERT solutions. 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. 

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.

Inverse Trigonometric Functions Chapter Summary - Class 12 NCERT Solutions

• The inverse trigonometric functions are the inverse functions of the trigonometric functions.

• The domain and the range of the trigonometric functions are converted into the range and domain of the inverse trigonometric functions.

• The domains and ranges (principal value branches) of inverse trigonometric function are given in the following table:

• $\sin ^{-1} x$ should not be confused with $(\sin x)^{-1}$. In fact $(\sin x)^{-1}-\frac{1}{\sin x}$ and similarly for other trigonometric functions,

• The value of an inverse trigonometric functions which lies in its principal value branch is called the principal value of that inverse trigonometric functions.

• For suitable values of domain, we have

Overview of Deleted Syllabus for CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Class 12 Maths NCERT Solutions Chapter 2 All Exercises

Conclusion

The NCERT Solutions for Class 12 Maths on inverse trigonometric functions provided by Vedantu are crucial for understanding this complex topic in a simple manner. It's important to focus on the concept of inverse trigonometric functions, their properties, and their applications in real-life scenarios. These solutions help students grasp the fundamental concepts and solve problems with ease. Previous year question papers typically include around 3-4 questions related to inverse trigonometric functions. Vedantu's NCERT Solutions offer a comprehensive understanding of inverse trigonometric functions, aiding students in scoring well and building a strong foundation in mathematics.

Other Study Materials of CBSE Class 12 Maths Chapter 2

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

Related Links for NCERT Class 12 Maths in Hindi

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