Answer
$\begin{align*}
\displaystyle \coth^{2}x-\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{2}x&=\displaystyle \frac{\cosh^{2}x}{\sinh^{2}x}-\frac{1}{\sinh^{2}x} \\
& =\displaystyle \frac{\cosh^{2}x-1}{\sinh^{2}x}\tag{ $\cosh^{2}x-\sinh^{2}x=1$ }\\
&=\displaystyle \frac{\sinh^{2}x}{\sinh^{2}x} \\
&=1 \end{align*}$
Work Step by Step
Definitions of the Hyperbolic Functions
$\displaystyle \sinh x=\frac{e^{x}-e^{-x}}{2} \qquad \mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}\ x=\displaystyle \frac{1}{\sinh x}=\frac{2}{e^{x}-e^{-x}},\quad x\neq 0$
$\displaystyle \cosh x=\frac{e^{x}+e^{-x}}{2} \qquad \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\ x=\displaystyle \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}$
$\displaystyle \tanh x=\frac{\sinh x}{\cosh x} =\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \qquad \coth x=\frac{1}{\tanh x}=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}},\quad x\neq 0$
And, we use $ \qquad \cosh^{2}x-\sinh^{2}x=1$
$(\displaystyle \frac{e^{2x}+2+e^{-2x}+4}{4}-\frac{e^{2x}-2+e^{-2x}+4}{4}=\frac{2+2}{4}=1)$
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$\begin{align*}
\displaystyle \coth^{2}x-\mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}^{2}x&=\displaystyle \frac{\cosh^{2}x}{\sinh^{2}x}-\frac{1}{\sinh^{2}x} \\
& =\displaystyle \frac{\cosh^{2}x-1}{\sinh^{2}x}\tag{ \cosh^{2}x-\sinh^{2}x=1 }\\
&=\displaystyle \frac{\sinh^{2}x}{\sinh^{2}x} \\
&=1 \end{align*}$
