physicscatalyst.com logo




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 
NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions EXERCISE 2.1

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
 

Also Read





Go back to Class 12 Main Page using below links
Class 12 Maths Class 12 Physics Class 12 Chemistry Class 12 Biology


Latest Updates
Synthetic Fibres and Plastics Class 8 Practice questions

Class 8 science chapter 5 extra questions and Answers

Mass Calculator

3 Fraction calculator

Garbage in Garbage out Extra Questions7