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}$
