#### Explain solution RD Sharma class 12  chapter Inverse Trigonometric Functions exercise 3.5 question 1 sub question (iv) maths

$\frac{-\pi}{2}$

Hints: $\operatorname{cosec}^{-1}$ is a function whose range of principal branch is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$

Thus, $\operatorname{cosec}^{-1}$ : R-(-1,1)→$\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$

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

Explanation: let $y = \left(2 \cos \frac{2 \pi}{3}\right)$

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

Range of function $\operatorname{cosec}^{-1}$ is $\left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ] \left \{ 0 \right \}$

Thus, $\operatorname{cosec}y = \left(2 \cos \frac{2 \pi}{3}\right)$

$\operatorname{cosec}y = 2 \times \frac{-1}{2}$                                                                                                                       $\left [\cos \frac{2\pi }{3}= \frac{-1}{2} \right ]$

$\operatorname{cosec}y =-1$

\begin{aligned} &\operatorname{cosec} y=\operatorname{cosec}\left(\frac{-\pi}{2}\right)\\ &y=\frac{-\pi}{2} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right], y \neq 0\\ &\text { Hence principal value of } \operatorname{cosec}^{-1}\left(2 \cos \frac{2 \pi}{3}\right) \text { is } \frac{-\pi}{2} \end{aligned}