#### Please solve RD Sharma class 12 Chapter Inverse Trigonometric Functions exercise 3.5 question 1 sub question (iii) maths textbook solution.

$\frac{-\pi}{3}$

Hints: The $\operatorname{cosec}^{-1}$ function is a function whose domain is R-(-1,1) and The principle value branch of $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$

Given: $\operatorname{cosec}^{-1} \left ( \frac{2}{\sqrt{3}} \right )$

Explanation: let $y=\operatorname{cosec}^{-1} \left ( \frac{2}{\sqrt{3}} \right )$

$\operatorname{cosec} y = \left ( \frac{2}{\sqrt{3}} \right )$

$\operatorname{cosec} y = \operatorname{cosec} \left ( \frac{\pi}{\sqrt{3}} \right )$                                                                                                                    $\left[\because \operatorname{cosec} \frac{\pi}{3}=\frac{2}{\sqrt{3}}\right]$

$\therefore y = \frac{\pi }{3}$

Hence the principal value of $\operatorname{cosec}^{-1} \left ( \frac{2}{\sqrt{3}} \right )$ is $\frac{\pi}{3}$