Provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise  3.7 question 5 sub question (iv)

Answer:    $-\frac{\pi }{6}$

Hint: The range of principal value of    $\operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
Given:  $\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{11 \pi}{6}\right)$

Explanation:

First we solve  $\operatorname{cosec}\left(\frac{11 \pi}{6}\right)$

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

As we know, $\operatorname{cosec}(2 \pi-\theta)=-\operatorname{cosec}(\theta)$

$\operatorname{cosec}\left(2 \pi-\frac{\pi}{6}\right)=-\operatorname{cosec}\left(\frac{\pi}{6}\right)$

As we know,  $\operatorname{cosec}\left(\frac{\pi}{6}\right)=2$

$-\operatorname{cosec}\left(\frac{\pi}{6}\right)=-2$

By substituting these values in $\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{11 \pi}{6}\right)$ we get,

$\operatorname{cosec}^{-1}(-2)$

Let,        $y=\operatorname{cosec}^{-1}(-2)$

\begin{aligned} &\operatorname{cosec} y=-2 \\ &-\operatorname{cosec} y=2 \\ &-\operatorname{cosec}\left(\frac{\pi}{6}\right)=2 \end{aligned}

As we know   $\operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta$

$-\operatorname{cosec} \frac{\pi}{6}=\operatorname{cosec}\left(-\frac{\pi}{6}\right)$

The range of principal value of    $\operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \text { and } \operatorname{cosec}\left(-\frac{\pi}{6}\right)=-2$

$\therefore \operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(\frac{11 \pi}{6}\right)\right) \mathrm{is}-\frac{\pi}{6}$