explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 5 sub question (ii) maths

Answer: $\frac{\pi }{4}$

Hint: The range of principal value of    $\operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
Given:  $\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{3 \pi}{4}\right)$

Explanation:

First we solve $\operatorname{cosec}\left(\frac{3 \pi}{4}\right)$

$\operatorname{cosec}\left(\frac{3 \pi}{4}\right)=\operatorname{cosec}\left(\pi-\frac{\pi}{4}\right)$

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

\begin{aligned} &\operatorname{cosec}\left(\pi-\frac{\pi}{4}\right)=\operatorname{cosec}\left(\frac{\pi}{4}\right) \\ &\operatorname{cosec}\left(\frac{\pi}{4}\right)=\sqrt{2} \end{aligned}

By substituting these values in $\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{3 \pi}{4}\right)$ we get,

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

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

\begin{aligned} &\operatorname{cosec} y=\sqrt{2} \\ &\operatorname{cosec}\left(\frac{\pi}{4}\right)=\sqrt{2} \end{aligned}

The range of the principal value of   $\operatorname{cosec}^{-1} \text { is }\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\} \text { and } \operatorname{cosec}\left(\frac{\pi}{4}\right)=\sqrt{2}$

$\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{\pi}{4}\right)=\frac{\pi}{4}$

Hence,  $\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{3 \pi}{4}\right)=\frac{\pi}{4}$