Answer
$\begin{align*}
\displaystyle \sinh 2x+\cosh 2x&=\displaystyle \frac{e^{2x}-e^{-2x}}{2}+\frac{e^{2x}+e^{-2x}}{2}\tag{ ... by definitions }\\
&=\displaystyle \frac{2e^{2x}}{2} \\
& =e^{2x} \end{align*}$
Work Step by Step
Definitions of the Hyperbolic Functions
$\displaystyle \sinh x=\frac{e^{x}-e^{-x}}{2} \qquad \mathrm{c}\mathrm{s}\mathrm{c}\mathrm{h}\ x=\displaystyle \frac{1}{\sinh x}=\frac{2}{e^{x}-e^{-x}},\quad x\neq 0$
$\displaystyle \cosh x=\frac{e^{x}+e^{-x}}{2} \qquad \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{h}\ x=\displaystyle \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}$
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$\begin{align*}
\displaystyle \sinh 2x+\cosh 2x&=\displaystyle \frac{e^{2x}-e^{-2x}}{2}+\frac{e^{2x}+e^{-2x}}{2}\tag{ ... by definitions }\\
&=\displaystyle \frac{2e^{2x}}{2} \\
& =e^{2x} \end{align*}$
So, $\qquad \sinh 2x+\cosh 2x=e^{2x}$
