#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 2 Subquestion (ii) Maths Textbook Solution

$\frac{\pi }{6}$
Hint:
$\sin^{-1}x= y$
$\sin y= x$
Given:
$Find\: principal\, value \, o\! f \sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \}$
Solution:
$\sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \}$
$= \sin^{-1}\left [ \cos \left ( \frac{\pi }{3} \right ) \right ] \; \; \; \; \; \; \; \; \left [ \because \left ( \frac{\sqrt{3}}{2} \right ) = \frac{\pi }{3}\right ]$
$= \left ( \frac{1}{2} \right )\; \; \; \; \; \; \; \; \left [ \because \cos \frac{\pi }{3}= \frac{1}{2} \right ]$

$Let , \left ( \frac{1}{2} \right )= y$
$\sin \sin y =\frac{1}{2}$
$Range\, o\! f \sin^{-1} is \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and \, \sin \sin \left ( \frac{\pi }{6} \right ) =\frac{1}{2}$
$Therefore, principal \, value\, o\! f \, \sin^{-1}\left \{ \cos \left ( \sin^{-1}\left ( \frac{\sqrt{3}}{2} \right ) \right ) \right \} is \, \frac{\pi }{6}.$