Show that $\sin^{-1}(\operatorname{cosec}x) = [2n + (-1)^n]\pi/2 + I(-1)^n \log(\cot x/2)$

Question

It is a question from hyperbolic and inverse hyperbolic functions

Answer 1

Apply the sine to both sides, use $\sin(x+k\pi)=(-1)^k\sin(x)=\sin((-1)^kx)$ to arrive at $$ \frac1{\sin x}=\sin\left(\frac\pi2+i\log(\cot\frac x2)\right)=\cosh(\log(\cot\frac x2))=\frac1{2}\left(e^{\log(\cot\frac x2)}+e^{-\log(\cot\frac x2)}\right) $$ so that in forward direction there are no contradictions. Now to go into the backwards direction you need to account for the branches of the inverse sine.