Answer
$\begin{align*}
2\displaystyle \sinh x\cosh x&=2(\displaystyle \frac{e^{x}-e^{-x}}{2})(\frac{e^{x}+e^{-x}}{2}) \tag{ numerator= diff. of squares }\\
&=\displaystyle \frac{e^{2x}-e^{-2x}}{2} \\
&=\sinh 2x \end{align*}$
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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$\begin{align*}
2\displaystyle \sinh x\cosh x&=2(\displaystyle \frac{e^{x}-e^{-x}}{2})(\frac{e^{x}+e^{-x}}{2}) \tag{ numerator= diff. of squares }\\
&=\displaystyle \frac{e^{2x}-e^{-2x}}{2} \\
&=\sinh 2x \end{align*}$
So, $\qquad 2\sinh x\cosh x=\sinh 2x$
