Prove that $\coth ^2(x)-1\equiv \mathrm{cosech}^2 (x)$
My attempts,
$\coth^2 (x)-1\equiv(\frac{e^x+e^{-x}}{e^x-e^{-x}})^2-1$
$\equiv \frac{e^{2x}+e^{-2x}+2}{e^{2x}+e^{-2x}-2}-1$
$\equiv \frac{e^{2x}+e^{-2x}+2}{e^{2x}+e^{-2x}-2}-\frac{e^{2x}+e^{-2x}-2}{e^{2x}+e^{-2x}-2}$
$\equiv \frac{e^{2x}+e^{-2x}+2-e^{2x}-e^{-2x}+2}{e^{2x}+e^{-2x}-2}$
$\equiv \frac{4}{e^{2x}+e^{-2x}-2}$
$\equiv (\frac{2}{e^x-e^{-x}})^2 $
$\equiv \mathrm{cosech}^2x$
Am I correct?