Answer
$1$
Work Step by Step
Definitions of the Hyperbolic Functions ...
$\displaystyle \tanh x=\frac{\sinh x}{\cosh x} =\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \qquad \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\ x=\displaystyle \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}$
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$\displaystyle \tanh^{2}x+\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}^{2}x=(\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}})^{2}+(\frac{2}{e^{x}+e^{-x}})^{2}$
$=\displaystyle \frac{e^{2x}-2+e^{-2x}+4}{(e^{x}+e^{-x})^{2}}$
$=\displaystyle \frac{e^{2x}+2+e^{-2x}}{e^{2x}+2+e^{-2x}}$
$=1$
