Chapter 6.9, Problem 62ES

Prove:

$\begin{array}{c}{\mathrm{sech}}^{-1}x={\cosh }^{-1}\left(1/x\right),0<x\le 1\\ {\coth }^{-1}x={\tanh }^{-1}\left(1/x\right),\left|x\right|>1\\ {\mathrm{csch}}^{-1}x={\sinh }^{-1}\left(1/x\right),x\ne 0\end{array}$