#### Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Functions exercise 3.5 question 3 sub question (iv)

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

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

Given: $\operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)$

Explanation:

$\operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)$

$\operatorname{cosec}^{-1}\left ( 2 \times \left ( \frac{-1}{\sqrt{3}} \right ) \right )$                                                                                                $\left [\because \tan \frac{11 \pi}{6}= \frac{-1}{\sqrt{3}} \right ]$

\begin{aligned} &\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right) \\ &\text { Let, } y=\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right) \\ &\qquad \operatorname{cosec} y=\left(\frac{-2}{\sqrt{3}}\right) \end{aligned}

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

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

$\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\frac{-\pi}{3} \quad \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$

Therefore, the principal value of $\operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)$ is $\frac{-\pi}{3}$