NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.1 - 2025-26

Exercise 2.1

1.  Find the principal value of ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ 

Ans: Let’s assume when  ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y$, Then $\sin y = \left( { - \dfrac{1}{2}} \right) =  - \sin \left( {\dfrac{\pi }{6}} \right) = \sin \left( { - \dfrac{\pi }{6}} \right)$.

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

Hence, the principal value of ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ is $ - \dfrac{\pi }{6}$.

2. Find the principal value of ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ 

Ans: Let’s consider, ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = y.$ Then $\cos y = \dfrac{{\sqrt 3 }}{2} = \cos \left( {\dfrac{\pi }{6}} \right)$

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2}$

Therefore, the principal value of ${\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ is $\dfrac{\pi }{6}$.

3. Find the principal value of ${\operatorname{cosec} ^{ - 1}}(2)$

Ans: Let’s consider,${\operatorname{cosec} ^{ - 1}}(2) = y$.Then, $\operatorname{cosec} {\text{y}} = 2 = \operatorname{cosec} \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\operatorname{cosec} ^{ - 1}}$ is$\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] - \{ 0\} $. 

Therefore, the principal value of ${\operatorname{cosec} ^{ - 1}}(2)$ is $\dfrac{\pi }{6}$.

4. Find the principal value of ${\tan ^{ - 1}}( - \sqrt 3 )$

Ans: Let’s consider ${\tan ^{ - 1}}( - \sqrt 3 ) = y$ Then, $\tan y =  - \sqrt 3  =  - \tan \dfrac{\pi }{3} = \tan \left( { - \dfrac{\pi }{3}} \right)$.

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and$\tan \left( { - \dfrac{\pi }{3}} \right)$ is $ - \sqrt 3 $

Hence, the principal value of ${\tan ^{ - 1}}( - \sqrt 3 )$ is $ - \dfrac{\pi }{3}$.

5. Find the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$

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

Then, $\cos y =  - \dfrac{1}{2} =  - \cos \left( {\dfrac{\pi }{3}} \right) = \cos \left( {\pi  - \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3}} \right)$.

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{{2\pi }}{3}} \right) =  - \dfrac{1}{2}$

Therefore, the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ is $\left( {\dfrac{{2\pi }}{3}} \right)$.

6. Find the principal value of ${\tan ^{ - 1}}( - 1)$

Ans: Let’s assume that ${\tan ^{ - 1}}( - 1) = {\text{y}}$.

Then, $\tan y =  - 1 =  - \tan \left( {\dfrac{\pi }{4}} \right) = \tan \left( { - \dfrac{\pi }{4}} \right)$.

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and$\tan \left( { - \dfrac{\pi }{4}} \right) =  - 1$

Therefore, the principal value of ${\tan ^{ - 1}}( - 1)$ is $ - \dfrac{\pi }{4}$.

7. Find the principal value of ${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right)$

Ans: Let’s consider ${\sec ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right) = y$. 

Then, $\sec y = \dfrac{2}{{\sqrt 3 }} = \sec \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\sec ^{ - 1}}$ is $[0,\pi ] - \left\{ {\dfrac{\pi }{2}} \right\}$ and $\sec \left( {\dfrac{\pi }{6}} \right) = \dfrac{2}{{\sqrt 3 }}$

8. Find the principal value of ${\cot ^{ - 1}}(\sqrt 3 )$

Ans: Let’s consider ${\cot ^{ - 1}}(\sqrt 3 ) = y$. Then $\cot y = \sqrt 3  = \cot \left( {\dfrac{\pi }{6}} \right)$.

As we know that the range of the principal value branch of ${\cot ^{ - 1}}$ is $(0,\pi )$ and $\cot \left( {\dfrac{\pi }{6}} \right) = \sqrt 3 $.

Hence, the principal value of ${\cot ^{ - 1}}(\sqrt 3 )$ is $\dfrac{\pi }{6}$.

9. Find the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$

Ans: Let’s ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) = y$.

Then $\cos y =  - \dfrac{1}{{\sqrt 2 }} =  - \cos \left( {\dfrac{\pi }{4}} \right) = \cos \left( {\pi  - \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{{3\pi }}{4}} \right)$.

As we know that the range of the principal value branch of ${\cos ^{ - 1}}$ is $[0,\pi ]$ and $\cos \left( {\dfrac{{3\pi }}{4}} \right) =  - \dfrac{1}{{\sqrt 2 }}$.

Therefore, the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$ is $\dfrac{{3\pi }}{4}$.

10. Find the principal value of ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 )$

Ans: Let’s consider, ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 ) = y$. Then, $\cos ecy =  - \sqrt 2  =  - \cos ec\left( {\dfrac{\pi }{4}} \right) = \cos ec\left( { - \dfrac{\pi }{4}} \right)$

As we know that the range of the principal value branch of ${\operatorname{cosec} ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] - \{ 0\} $ and $\operatorname{cosec} \left( { - \dfrac{\pi }{4}} \right) =  - \sqrt 2 $

Therefore, the principal value of ${\operatorname{cosec} ^{ - 1}}( - \sqrt 2 )$ is $ - \dfrac{\pi }{4}$.

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

Ans: Let’s consider ${\tan ^{ - 1}}(1) = x$. Then, $\tan x = 1 = \tan \left( {\dfrac{\pi }{4}} \right)$.

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

Let’s assume,${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = y$.

Then, $\cos y =  - \dfrac{1}{2} =  - \cos \left( {\dfrac{\pi }{3}} \right) = \cos \left( {\pi  - \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3}} \right)$.

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

Let’s again assume that ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = z$.

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

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

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

$ = \dfrac{\pi }{4} + \dfrac{{2\pi }}{3} - \dfrac{\pi }{6}$

$ = \dfrac{{3\pi  + 8\pi  - 2\pi }}{{12}} = \dfrac{{9\pi }}{{12}} = \dfrac{{3\pi }}{4}$

12. Find the value of ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$

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

Then, $\cos x = \dfrac{1}{2} = \cos \left( {\dfrac{\pi }{3}} \right)$.

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

Let’s assume ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = y.$

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

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

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

13. If ${\sin ^{ - 1}}x = y$, then

(A) $0 \leqslant {\text{y}} \leqslant \pi $

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

(C) $0 < y < \pi $

(D) $ - \dfrac{\pi }{2} < y < \dfrac{\pi }{2}$

Ans: It is given that ${\sin ^{ - 1}}x = y$. As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$. 

Therefore, $ - \dfrac{\pi }{2} \leqslant y \leqslant \dfrac{\pi }{2}$.

Hence option (B) is correct.

14. ${\tan ^{ - 1}}\sqrt 3  - {\sec ^{ - 1}}( - 2)$ is equal to

(A) $\pi $

(B) $ - \pi /3$

(C) $\pi /3$

(D) $2\pi /3$

Ans: Let’s consider, ${\tan ^{ - 1}}\sqrt 3  = x.$.

Then, $\tan x = \sqrt 3  = \tan \dfrac{\pi }{3}$

As we know that the range of the principal value branch of ${\tan ^{ - 1}}$ is $\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$. $\therefore {\tan ^{ - 1}}\sqrt 3  = \dfrac{\pi }{3}$

Let assume, ${\sec ^{ - 1}}( - 2) = y$.

Then, $\sec y =  - 2 =  - \sec \left( {\dfrac{\pi }{3}} \right) = \sec \left( {\pi  - \dfrac{\pi }{3}} \right) = \sec \dfrac{{2\pi }}{3}$.

As we know that the range of the principal value branch of sec $^1$ is $[0,\pi ] - \left\{ {\dfrac{\pi }{2}} \right\}$. $\therefore {\sec ^{ - 1}}( - 2) = \dfrac{{2\pi }}{3}$

Thus, ${\tan ^{ - 1}}(\sqrt 3 ) - {\sec ^{ - 1}}( - 2)$

$ = \dfrac{\pi }{3} - \dfrac{{2\pi }}{3} =  - \dfrac{\pi }{3}$

Conclusion

NCERT Solutions for Class 12 Maths Chapter 2 - Inverse Trigonometric Functions by Vedantu offer a detailed guide to mastering key topics such as definitions, domains, ranges, principal values, and properties. Focus on understanding principal values and practicing properties to simplify expressions. These solutions provide a clear explanation of each problem, aiding in conceptual clarity. Vedantu ensures that students develop a solid grasp of the subject matter, making it easier to tackle complex problems. Additionally, the solutions are designed to align with the latest syllabus, ensuring thorough preparation for exams.

Class 12 Maths Chapter Inverse Trigonometric Functions: Exercises Breakdown

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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.