#### Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 5 sub question (iii) maths textbook solution

Answer:    $-\frac{\pi }{5}$

Hint: The range of principal value of    $\operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
Given:    $\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{6 \pi}{5}\right)$

Explanation:

First we solve $\operatorname{cosec}\left(\frac{6 \pi}{5}\right)$

$\operatorname{cosec}\left(\frac{6 \pi}{5}\right)=\operatorname{cosec}\left(\pi+\frac{\pi}{5}\right)$

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

Then,     $\operatorname{cosec}\left(\pi+\frac{\pi}{5}\right)=-\operatorname{cosec}\left(\frac{\pi}{5}\right)$

$-\operatorname{cosec}\left(\frac{\pi}{5}\right)=\operatorname{cosec}\left(-\frac{\pi}{5}\right)$

As         $\operatorname{cosec}(-x)=-\operatorname{cosec} x$

By substituting this value in $\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{6 \pi}{5}\right)$ we get,

$\operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(-\frac{\pi}{5}\right)\right)$

As we know,   $\operatorname{cosec}^{-1}(\operatorname{cosec} x)=x \text { when } x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$

$\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{6 \pi}{5}\right)=-\frac{\pi}{5}$