NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.2 - 2025-26

Refer to page 10 for Exercise 2.2 in the PDF. 

1. Prove $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right),x \in \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]$

Ans: To Prove $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$, where $x \in \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]$

Let $x = \sin \theta .$ Then, ${\sin ^{ - 1}}x = \theta $.

We have given that,

R.H.S ${\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) = {\sin ^{ - 1}}\left( {3\sin \theta  - 4{{\sin }^3}\theta } \right)$

$ = {\sin ^{ - 1}}(\sin 3\theta )$

$ = 3\theta $

$ = 3{\sin ^{ - 1}}x = L.H.S$

Hence proved.

2. Prove $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\dfrac{1}{2},1} \right]$

Ans: To Prove $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\dfrac{1}{2},1} \right]$

Let $x = \cos \theta $. Then, ${\cos ^{ - 1}}x = \theta $

We have given that,

R.H S =${\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)$

$ = {\cos ^{ - 1}}\left( {4{{\cos }^3}\theta  - 3\cos \theta } \right)$

$ = {\cos ^{ - 1}}(\cos 3\theta )$

$ = 3\theta $

$ = 3{\cos ^{ - 1}}x = L.H.S$

Hence proved.

3. Write the function in the simplest form: ${\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x},x \ne 0$

Ans: ${\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x}$

By putting $x = \tan \theta  \Rightarrow \theta  = {\tan ^{ - 1}}x$

$\therefore {\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x} = {\tan ^{ - 1}}\dfrac{{\sqrt {1 + {{\tan }^2}\theta }  - 1}}{{\tan \theta }}$

$ = {\tan ^{ - 1}}\left( {\dfrac{{\sec \theta  - 1}}{{\tan \theta }}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 - \cos \theta }}{{\sin \theta }}} \right)$

${\tan ^{ - 1}}\left( {\dfrac{{2{{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\tan \dfrac{\theta }{2}} \right) = \dfrac{\theta }{2} = \dfrac{1}{2}{\tan ^{ - 1}}x$

4. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right),x < \pi $

Ans: Given that,${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right),x < \pi $

${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right)$

$ = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}}} } \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2}} \right)$

$ = \dfrac{x}{2}$

5. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right),0 < x < \pi $

Ans: We have given that, ${\tan ^{ - 1}}\left( {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{1 + \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)$

$ = {\tan ^{ - 1}}(1) - {\tan ^{ - 1}}(\tan x)\quad \left[ {\because \quad \dfrac{{ - y}}{{x - xy}} = {{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right]$

$ = \dfrac{\pi }{4} - x$

6. Write the function in the simplest form: ${\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }},|x| < a$

Ans: We have given that,${\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$

Let’s consider, $x = a\sin \theta  \Rightarrow \dfrac{x}{a} = \sin \theta  \Rightarrow {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

$\therefore {\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\sqrt {1 - {{\sin }^2}\theta } }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\cos \theta }}} \right)$

$ = {\tan ^{ - 1}}(\tan \theta ) = \theta  = {\sin ^{ - 1}}\dfrac{x}{a}$

7. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),a > 0;\dfrac{{ - a}}{{\sqrt 3 }} \leqslant x \leqslant \dfrac{a}{{\sqrt 3 }}$

Ans: Consider, ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

Let’s consider, $x = a\tan \theta  \Rightarrow \dfrac{x}{a} = \tan \theta  \Rightarrow \theta  = {\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{3{a^2} \cdot a\tan \theta  - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3a \cdot {a^2}{{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{3{a^3}\tan \theta  - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3{a^3}{{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}(\tan 3\theta )$

$ = 3\theta $

$ = 3{\tan ^{ - 1}}\dfrac{x}{a}$

8. Find the value of ${\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\dfrac{1}{2}} \right)} \right]$

Ans: Let’s consider, ${\sin ^{ - 1}}\dfrac{1}{2} = x$

Then $\sin x = \dfrac{1}{2} = \sin \left( {\dfrac{\pi }{6}} \right)$

$\therefore {\sin ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{6}$

$\therefore {\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\dfrac{1}{2}} \right)} \right]$

$ = {\tan ^{ - 1}}\left[ {2\cos \left( {2 \times \dfrac{\pi }{6}} \right)} \right]$

$ = {\tan ^{ - 1}}\left[ {2\cos \dfrac{\pi }{3}} \right]$

$ = {\tan ^{ - 1}}\left[ {2 \times \dfrac{1}{2}} \right]$

$ = {\tan ^{ - 1}}1 = \dfrac{\pi }{4}$

9. Find the value of $\tan \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right],|x| < 1,y > 0$ and $xy < 1$

Ans: Let consider, $x = \tan \theta $.

Then, $\theta  = {\tan ^{ - 1}}x$.

$\therefore {\sin ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} = {\sin ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\sin ^{ - 1}}(\sin 2\theta )$

$ = 2\theta $

$ = 2{\tan ^{ - 1}}x$

Let’s assume, $y = \tan \theta .$ Then, $\theta  = {\tan ^{ - 1}}y$.

$\therefore {\cos ^{ - 1}}\left( {\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right)$

$ = {\cos ^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\cos ^{ - 1}}(\cos 2\theta )$

$ = 2\theta  = 2{\tan ^{ - 1}}y$

$\therefore \tan \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right)} \right]$

$ = \tan \dfrac{1}{2}\left[ {2{{\tan }^{ - 1}}x + 2{{\tan }^{ - 1}}y} \right]$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$ = \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)} \right]$

$ = \dfrac{{x + y}}{{1 - xy}}$

10. Find the values of ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

Ans: Let’s consider, ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

As we know that ${\sin ^{ - 1}}(\sin x) = x$

If $x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$, which is the principal value branch of ${\sin ^{ - 1}}x$.

Here, $\dfrac{{2\pi }}{3} \notin \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$

Now, ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$ can be written as:

${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

$ = {\sin ^{ - 1}}\left[ {\sin \left( {\pi  - \dfrac{{2\pi }}{3}} \right)} \right]$

$ = {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{3}} \right)$, where $\dfrac{\pi }{3} \in \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$

$\therefore {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right) = {\sin ^{ - 1}}\left[ {\sin \dfrac{\pi }{3}} \right] = \dfrac{\pi }{3}$

11. Find the values of ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$

Ans: Let’s Consider, ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$

As we know that ${\tan ^{ - 1}}(\tan x) = x$

If $x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$, which is the principal value branch of ${\tan ^{ - 1}}x$. Here, $\dfrac{{3\pi }}{4} \notin \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.

Now, ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$ can be written as:

${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right) = {\tan ^{ - 1}}\left[ { - \tan \left( {\dfrac{{ - 3\pi }}{4}} \right)} \right] = {\tan ^{ - 1}}\left[ { - \tan \left( {\pi  - \dfrac{\pi }{4}} \right)} \right]$

${\tan ^{ - 1}}\left( { - \tan \dfrac{\pi }{4}} \right) = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right]$ where $ - \dfrac{\pi }{4} \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$

$\therefore {\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right) = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right] = \dfrac{{ - \pi }}{4}$

12. Find the values of $\tan \left( {{{\sin }^{ - 1}}\dfrac{3}{5} + {{\cot }^{ - 1}}\dfrac{3}{2}} \right)$

Ans: Let’s consider, ${\sin ^{ - 1}}\dfrac{3}{5} = x$.

Then, $\sin x = \dfrac{3}{5}$

$ \Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x}  = \dfrac{4}{5}$

$ \Rightarrow \sec x = \dfrac{5}{4}$

$\therefore \tan x = \sqrt {{{\sec }^2}x - 1}  = \sqrt {\dfrac{{25}}{{16}} - 1}  = \dfrac{3}{4}$

$\therefore x = {\tan ^{ - 1}}\dfrac{3}{4}$

$\therefore {\sin ^{ - 1}}\dfrac{3}{5} = {\tan ^{ - 1}}\dfrac{3}{4}\quad  \ldots (i)$

Therefore, $\tan \left( {{{\sin }^{ - 1}}\dfrac{3}{5} + {{\cot }^{ - 1}}\dfrac{3}{2}} \right)$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{3}{4} + {{\tan }^{ - 1}}\dfrac{2}{3}} \right)\quad $ [Using (i) and (ii)]

$ = \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\dfrac{3}{4} + \dfrac{2}{3}}}{{1 - \dfrac{3}{4} \cdot \dfrac{2}{3}}}} \right)} \right]$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{9 + 8}}{{12 - 6}}} \right)$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{17}}{6}} \right) = \dfrac{{17}}{6}$

13. Find the values of ${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right)$ is equal to

(A) $\dfrac{{7\pi }}{6}$

(B) $\dfrac{{5\pi }}{6}$

(C) $\dfrac{\pi }{3}$

(D) $\dfrac{\pi }{6}$

Ans: We know that ${\cos ^{ - 1}}(\cos x) = x$ if $x \in [0,\pi ]$, which is the principal value branch of ${\cos ^{ - 1}}x$. Here, $\dfrac{{7\pi }}{6} \notin [0,\pi ]$.

Now, ${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right)$ can be written as:

${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{ - 7\pi }}{6}} \right) = {\cos ^{ - 1}}\left[ {\cos \left( {2\pi  - \dfrac{{7\pi }}{6}} \right)} \right]\quad [\because $

$ + x) = \cos x]$

$\therefore {\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{6}} \right) = \dfrac{{5\pi }}{6}$

The correct answer is ${\text{B}}$.

14. Find the values of $\sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right)$ is equal to

(A) $\dfrac{1}{2}$

(B) $\dfrac{1}{3}$

(C) $\dfrac{1}{4}$

(D) 1

Ans: Let’s consider, ${\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = x$.

Then, $\sin x = \dfrac{{ - 1}}{2} =  - \sin \dfrac{\pi }{6} = \sin \left( {\dfrac{{ - \pi }}{6}} \right)$.

As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$. ${\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = \dfrac{\pi }{6}$

$\therefore \sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right) = \sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6}} \right) = \sin \left( {\dfrac{{3\pi }}{6}} \right) = \sin \left( {\dfrac{\pi }{2}} \right) = 1$

The correct answer is ${\text{D}}$.

15. $\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$ is equal to

(A) $\pi$

(B) $-\dfrac{\pi}{2}$

(C) 0

(D) $2 \sqrt{3}$

Ans:

Suppose that,

$A=\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$

As we know that,

$\cot ^{-1}(-x)=\pi-\cot ^{-1}(x)$

So,

$\cot ^{-1}(-\sqrt{3})=\pi-\cot ^{-1}(\sqrt{3})$

Put the value in the equation (i),

$A=\tan ^{-1}(\sqrt{3})-\pi+\cot ^{-1}(\sqrt{3})$

As we know that,

$\cot ^{-1} x=\dfrac{\pi}{2}-\tan ^{-1} x$

So,

$\cot ^{-1}(\sqrt{3})=\dfrac{\pi}{2}-\tan ^{-1}(\sqrt{3})$

Put the value in the equation (ii),

$A=\tan ^{-1}(\sqrt{3})-\pi+\dfrac{\pi}{2}-\tan ^{-1}(\sqrt{3}) $

$A=-\dfrac{\pi}{2}$

Hence, the correct option is B.

Conclusion

NCERT Solutions for Maths Chapter 2 Inverse Trigonometric Functions, Ex2.2 class 12 by Vedantu provides clear explanations and step-by-step solutions for understanding the properties and applications of inverse trigonometric functions. This exercise emphasizes the importance of knowing the principal value ranges for inverse functions, as well as solving equations involving these functions. Students should focus on mastering these ranges and practicing how to simplify and solve related equations accurately. Previous exam papers typically include 2-3 questions from this topic, highlighting its significance in understanding the broader subject of trigonometry. Vedantu’s solutions are designed to build a strong foundation, ensuring students can handle these concepts confidently in exams.

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