Answer
See the proof below.
Work Step by Step
Using the definition of $\cosh x$, we have the following
\begin{aligned} \cosh (x+y) &=\frac{e^{x+y}+e^{-(x+y)}}{2}=\frac{2 e^{x+y}+2 e^{-(x+y)}}{4} \\ &=\frac{e^{x+y}+e^{-x+y}+e^{x-y}+e^{-(x+y)}}{4}+\frac{e^{x+y}-e^{-x+y}-e^{x-y}+e^{-(x+y)}}{4} \\ &=\left(\frac{e^{x}+e^{-x}}{2}\right)\left(\frac{e^{y}+e^{-y}}{2}\right)+\left(\frac{e^{x}-e^{-x}}{2}\right)\left(\frac{e^{y}-e^{-y}}{2}\right) \\ &=\cosh x \cosh y+\sinh x \sinh y \end{aligned}
which completes the proof.
