Chapter 2: Inverse Trigonometric Functions EXERCISE 2.1

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
 

Also Read

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• Notes
  • Inverse Trigonometric Functions
  • Properties of Inverse Trigonometric Functions
  NCERT Solutions & Assignments
  • NCERT Solution for Inverse Trigonometric Functions Exercise 2.1
  • NCERT Solution for Inverse Trigonometric Functions Exercise 2.2
  • Inverse Trigonometric Functions Extra Questions for Class 12

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