#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 3 Subquestion (ii) Maths Textbook Solution.

Answer: $\frac{\pi }{4}$
Hints:
First we will find the principal value of $\sin^{-1}\frac{\sqrt{3}}2{}$
The principal value branch of function $\sin^{-1} is \left [ \frac{-\pi }{2} ,\frac{\pi }{2}\right ]$
The principal value branch of function $\cot ^{-1} is (0, \pi )$
Given: $cot^{-1}\left \{ 2 \cos\left ( \sin^{-1}\frac{\sqrt{3}}{2} \right ) \right \}$
Solution:
$cot^{-1}\left \{ 2 \cos\left ( \sin^{-1}\frac{\sqrt{3}}{2} \right ) \right \}$                        $\cdot \cdot \cdot (i)$
Let us first solve for   $\sin^{-1}\frac{\sqrt{3}}2{}$
Let   $x= \sin^{-1}\frac{\sqrt{3}}2{}$             $\cdot \cdot \cdot (ii)$
$\sin \: x = \frac{\sqrt{3}}2{}$
$\sin\: x = \sin \frac{\pi }{3}$
$x = \frac{\pi }{3}$

$\sin^{-1}\frac{\sqrt{3}}{2} = \frac{\pi }{3}$                                                                  [from equation (ii)]
The principal value branch of $\sin^{-1}$ is  $\left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]$
$\because \sin^{-1}\frac{\sqrt{3}}{2}= \frac{\pi }{3} \: \epsilon \: \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ]$
$\because$ Principal value of $\sin^{-1}\frac{\sqrt{3}}{2}$  is $\frac{\pi }{3}$                                                                 $\cdot \cdot \cdot (iii)$
Now from equation (i)
$cot^{-1}\left \{ 2 \cos\left ( \sin^{-1}\frac{\sqrt{3}}{2} \right ) \right \}$
$= \cot^{-1}\left \{ 2 \cos\left (\frac{\pi }{3} \right ) \right \}$                                                                       [from equation (iii)]
$= \cot^{-1} \left ( 2\times \frac{1}{2} \right )$
$=\ cot^{-1}\left ( 1 \right )$                                                                                                                   $\cdot \cdot \cdot (iv)$
Let $y=\ cot^{-1}\left ( 1 \right )$                                                                                                      $\cdot \cdot \cdot (v)$
$\cot y = 1$

$\cot\: y = \cot \frac{\pi }{4}$
$y = \frac{\pi }{4}$                                                                                                           [from v]
$\cot^{-1}\left ( 1 \right ) =\frac{\pi }{4}\: \epsilon \: (0, \pi )$
From equation (iv)
$cot^{-1}\left \{ 2 \cos\left ( \sin^{-1}\frac{\sqrt{3}}{2} \right ) \right \}$
$= \cot^{-1}\left ( 1 \right )$
$= \frac{\pi }{4}$