Answer
$ \cosh(x) + \sinh(x) =e^{x}$
Work Step by Step
Since, we have $\cosh(x) = \dfrac{e^{x} + e^{-x}}{2}$ and
$\sinh(x) = \dfrac{e^{x} - e^{-x}}{2}$
Thus, $ \cosh(x) + \sinh(x) = \dfrac{e^{x} + e^{-x}}{2}+\dfrac{e^{x} - e^{-x}}{2} $
or, $ = \dfrac{2e^{x}}{2}$
or, $ =e^{x}$
