Doubt in Inverse Trigonometric Function Problem

Question

Question: $\DeclareMathOperator{\cosec}{cosec}$

Let domain and range of $\cosec^{-1}(x)$ be $(-\infty,-1]\bigcup[1,\infty)$ and $[\frac{\pi}{2},\frac{3\pi}{2}]-\{\pi\}$. If $x\in(-\frac{\pi}{2},\frac{\pi}{2})$, then $\cosec^{-1}(\cosec(x))$ is

(A) $2\pi-x$

(B) $\pi+x$

(C) $\pi-x$

(D) $-\pi-x$

When the domain and range are principal, I used to convert the given domain or range to the principal by adding or subtracting $n\pi$. But, I tried here the same thing but I got the answer as D which is incorrect. Kindly help me solve this problem. I don't know any other method. Please explain.

Answer 1

We must have $y \in \bigg[\frac\pi2,\frac{3\pi}2\bigg] -\{\pi\}$ to obtain $\csc^{-1}(\csc y) = y$

Now $-\frac\pi2\lt x\lt \frac\pi2 \implies -\frac\pi2\lt- x\lt \frac\pi2 \implies \color{red}{\frac\pi2<-x+\pi<\frac{3\pi}2} \ x\ne0$

As, $\csc(\pi-x) = \csc(x)$

$$\csc^{-1}(\csc(x)) = \csc^{-1}(\csc(\pi-x)) = \pi-x \ , \pi-x\in\bigg(\frac\pi2,\frac{3\pi}2\bigg)-\{\pi\}$$