Answer
$\sinh x\cosh y+\cosh x\sinh y=\sinh(x+y)$
(please see details in step-by=step)
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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$\begin{align*}
\displaystyle \sinh x\cosh y&=(\displaystyle \frac{e^{x}-e^{-x}}{2})(\frac{e^{y}+e^{-y}}{2})\\
&=\displaystyle \frac{e^{x+y}-e^{-x+y}+e^{x-y}-e^{-x-y)}}{4} \\
&=\displaystyle \frac{(e^{x+y}-e^{-(x+y)})-e^{-x+y}+e^{x-y}}{4} \\\\\\
\displaystyle \cosh x\sinh y&=(\displaystyle \frac{e^{x}+e^{-x}}{2})(\frac{e^{y}-e^{-y}}{2}) \\
&=\displaystyle \frac{e^{x+y}+e^{-x+y}-e^{x-y}-e^{-(x+y)}}{4} \\
& =\displaystyle \frac{(e^{x+y}-e^{-(x+y)})+e^{-x+y}-e^{x-y}}{4}\\\\\\
\displaystyle \sinh x\cosh y+\cosh x\sinh y& =\frac{2(e^{x+y}-e^{-(x+y)})}{4}\\
& =\displaystyle \frac{e^{x+y}-e^{-(x+y)}}{2} \\
& =\sinh(x+y) \end{align*}$
