 Chapter 2: Inverse Trigonometric Functions EXERCISE 2.1

In this page we have NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions for EXERCISE 2.1 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Formula's used
Principle Value branch of Trigonometric Inverse Functions along with domain and range Find the principal values of the following

Question 1. $\sin^{-1}\left ( -\frac{1}{2} \right )$
Solution
Let $\sin^{-1}\left ( -\frac{1}{2} \right )$ = x, then
$\sin {x} = -\frac{1}{2} = - \sin\frac{ \pi}{6} = \sin ( - \frac{ \pi}{6} )$
We know from above table that
The principal value branch range for sin-1 is $\left [ -\frac{ \pi}{2}, \frac{ \pi}{2} \right ]$ and $\sin ( -\frac{ \pi}{6} ) = - \frac{1}{2}$
Therefore principal value for $\sin^{-1}\left ( -\frac{1}{2} \right ) \; is \; - \frac{ \pi}{6}$

Question 2. $\cos^{-1}\left ( - \frac{\sqrt{3}}{2} \right )$
Solution
Let $\cos^{-1}\left ( - \frac{\sqrt{3}}{2} \right )$ = x, then
$\cos x = \frac{\sqrt{3}}{2} = \cos (\frac{\pi}{6})$ We know from above table that
The principal value branch range for cos-1 is $\left [ 0 , \pi \right ]$ and $\cos (\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$
Therefore, principal value for $\cos^{-1}\left ( - \frac{\sqrt{3}}{2} \right ) \; is \; \frac{\pi}{6}$

Question 3. cosec-1 (2)
Solution
Let cosec-1 (2) = x. Then, cosec x = 2 = cosec $(\frac{\pi}{6} )$
We know from above table that
The principal value branch range for cosec-1 is $\left [ -\frac{\pi}{2}, \frac{\pi}{2}\right ] - {0}$ and cosec$(\frac{\pi}{6} )$ = 2
Therefore, principal value for cosec-1 (2) is $\frac{\pi}{6}$

Question 4. $\tan^{-1} \left ( - \sqrt{3} \right )$
Solution

Let $\tan^{-1} \left ( - \sqrt{3} \right ) = x$
Then, $\tan = - \sqrt{3} = - \tan \frac{\pi}{3} \tan (- \frac{\pi}{3})$
We know from above table that
The principal value branch range for $\tan^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ] \; and \; \tan\left ( -\frac{\pi}{3} \right ) = -\sqrt{3}$
Therefore, principal value for $\tan^{-1} \left ( - \sqrt{3} \right ) \; is \; -\frac{\pi}{3}$

Question 5. $\cos^{-1}\left ( -\frac{1}{2} \right )$
Solution
Let $\cos^{-1}\left ( -\frac{1}{2} \right )$ = x,
Then $\cos x = -\frac{1}{2} = -cos \frac{\pi}{3} = \cos ( \pi - \frac{\pi}{3} ) = \cos( \frac{2 \pi}{3} )$
We know from above table that
The principal value branch range for $\cos ^{-1} \; is \; \left [ 0 , \pi \right ] \; and \; \cos \left ( \frac{2 \pi}{3} \right ) = - \frac{1}{2}$
Therefore, principal value for $\cos^{-1}\left ( -\frac{1}{2} \right ) \; is \; \frac{2 \pi}{3}$

Question 6. $\tan^{-1} (-1)$
Solution

Let $\tan^{-1} (-1) = x$,
Then, tan x = -1 = $-\tan ( \frac{\pi}{4} ) = \tan ( - \frac{\pi}{4} )$
We know from above table that
The principal value branch range for $\tan^{-1} \; is \; \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right ) \; and \; \tan ( - \frac{\pi}{4} ) = -1$
Therefore, principal value for $\tan^{-1} (-1) \; is \; -\frac{\pi}{4}$

Question 7. $\sec^{-1} \left ( \frac{2}{\sqrt{3}} \right )$
Solution

Let $\sec^{-1} \left ( \frac{2}{\sqrt{3}} \right ) = x$,
Then $\sec x = \frac{2}{\sqrt{3}} = \sec (\frac{\pi}{6})$
We know from above table that
The principal value branch range for $\sec^{-1} \; is \; \left [ 0 , \pi \right ] - \left \{ \frac{ \pi }{2} \right \} \; and \; \sec (\frac{\pi}{6}) = \frac{2}{\sqrt{3}}$
Therefore, principal value for $\sec^{-1} \left ( \frac{2}{\sqrt{3}} \right ) \; is ; \frac{\pi}{6}$

Question 8. $\cot^{-1} \sqrt{3}$
Solution
Let $\cot^{-1} \sqrt{3} = x$,
Then $\cot x = \sqrt {3} = \cot \left ( \frac{\pi}{6} \right )$
We know from above table that
The principal value branch range for cot-1 is $( 0 , \pi )$ and $\cot \left ( \frac{\pi}{6} \right ) = \sqrt{3}$
Therefore, principal value for $\cot^{-1} \sqrt{3} = \frac{\pi}{6}$

Question 9. Find principal value for $\cos^{-1} \left ( - \frac{1}{\sqrt{2}} \right )$
Solution
Let $\cos^{-1} \left ( - \frac{1}{\sqrt{2}} \right ) = a$
Then $\cos x = \frac{-1}{\sqrt{2}} = - \cos \left ( \frac{\pi}{4} \right ) = \cos \left ( \pi - \frac{\pi}{4} \right ) = \cos \left ( \frac{3 \pi}{4} \right )$
We know from above table that
The principal value branch range for cos-1 is $[0 , \pi] \; and \; \cos \left ( \frac{3 \pi}{4} \right ) = -\frac{1}{\sqrt{2}}$
Therefore, principal value for $\cos^{-1} \left ( - \frac{1}{\sqrt{2}} \right ) \; is \; \frac{3 \pi }{4}$

Question 10. Find principal value for cosec-1 $\left ( -\sqrt{2} \right )$
Solution
Let cosec-1$\left ( -\sqrt{2} \right )$ = x, Then
cosec x = $-\sqrt{2}$ = -cosec$\left ( \frac{\pi}{4}\right )$ = cosec $\left ( -\frac{\pi}{4}\right )$
We know,
The principal value branch range for cosec-1 is $\left [ -\frac{\pi}{2} , \frac{\pi}{2} \right ] - \left \{ 0 \right \}$ and cosec$\frac{-\pi}{4} = -\sqrt{2}$
Therefore, principal value for cosec-1 $\left ( -\sqrt{2} \right ) \; is \; -\frac{\pi}{4}$

Find the values of the following

Question 11. Solve $\tan ^{-1}(1) + \cos^{-1} \left ( -\frac{1}{2} \right ) + \sin ^{-1}\left ( -\frac{1}{2} \right )$
Solution

Let $\tan ^{-1}(1) = x$, then
$\tan x = 1 = \tan \frac{\pi}{4}$ We know from above table that
The principal value branch range for $\tan ^{-1} \; is \; \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
$\tan ^{-1}(1) = \frac{\pi}{4}$
Let $\cos^{-1} \left ( -\frac{1}{2} \right ) = y$, then
$\cos y = -\frac{1}{2} = -\cos \frac {\pi}{3} = \cos\left ( \pi - \frac{\pi}{3} \right ) = \cos\left ( \frac{2\pi}{3} \right )$
We know from above table that
The principal value branch range for cos-1 is $[0 , \pi]$
$\cos ^{-1} \left ( -\frac{1}{2} \right ) = \frac{2 \pi }{3}$
Let $\sin^{-1}\left ( -\frac{1}{2} \right ) = z$, then
$\sin z = - \frac{1}{2} = - \sin \frac{\pi}{6} = \sin \left ( -\frac{\pi}{6} \right )$
We know from above table that
The principal value branch range for $\sin ^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$
$\sin^{-1} \left ( -\frac{1}{2} \right ) = - \frac{\pi}{6}$ Now
$\tan ^{-1}(1) + \cos^{-1} \left ( -\frac{1}{2} \right ) + \sin ^{-1}\left ( -\frac{1}{2} \right )$ $= \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi}{4}$
Question 12. Solve $\cos ^{-1} \left ( \frac{1}{2} \right ) + 2 \sin ^{-1} \left ( \frac{1}{2} \right )$
Solution
Let $\cos ^{-1} \left ( \frac{1}{2} \right ) = x$, then $\cos x = \frac{1}{2} = \cos \frac{\pi}{3}$
We know from above table that
The principal value branch range for cos-1 is $\left [0 , \pi \right ]$
$\cos ^{-1} \left ( \frac{1}{2} \right ) = \frac{\pi}{3}$
Let $\sin ^{-1} \left (- \frac{1}{2} \right ) = y$, then $\sin y = \frac{1}{2} = \sin \frac{\pi}{6}$
We know,
The principal value branch range for $\sin ^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$
$\sin ^{-1} \left ( \frac{1}{2} \right ) = \frac{\pi}{6}$
Now,
$\cos ^{-1} \left ( \frac{1}{2} \right ) + 2 \sin ^{-1} \left ( \frac{1}{2} \right )$ $= \frac{\pi}{3} + 2 \times \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$

Multiple Choice Questions

Question 13. If sin-1 a = b, then
(i) $0 \leq b \leq \pi$
(ii) $-\frac{\pi}{2} \leq b \leq \frac{\pi}{2}$
(iii) $0 < b < \pi$
(iv) $-\frac{\pi}{2} < b < \frac{\pi}{2}$
Solution
Given sin-1 a = b
We know from above table that
The principal value branch range for $\sin ^{-1} \; is \; \left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$
Therefore, $-\frac{\pi}{2} \leq b \leq \frac{\pi}{2}$

Question 14. The value of $\tan ^{-1} \sqrt{3} - \sec ^{-1}(-2)$ is
(i) $\pi$
(ii) $- \frac{\pi}{3}$
(iii)$\frac{\pi}{3}$
(iv) $\frac{2 \pi}{3}$
Solution
Let $\tan ^{-1} \sqrt{3} = x$, then
$\tan x = \sqrt{3} = \tan \frac{\pi}{3}$
We know from above table that
The principal value branch range for $\tan ^{-1} \; is \; \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
$\tan ^{-1}\sqrt{3} = \frac{\pi}{3}$
Let sec-1(-2) = y, then
sec y = -2 = $- \sec \frac{\pi}{3} = \sec \left ( \pi - \frac{\pi}{3} \right ) = \sec \left ( \frac{2 \pi}{3} \right )$
We know from above table that
The principal value branch range for sec-1 is $[0 , \pi] - \left \{ \frac{\pi}{2} \right \}$
$\sec ^{-1}(-2) = \frac{2 \pi}{3}$
Now,
$\tan ^{-1} \sqrt{3} - \sec ^{-1}(-2) = \frac{\pi}{3} - \frac{2 \pi}{3} = - \frac{\pi}{3}$
Hence option (ii) is correct