Establish the identity $\cosh ^{2}x-\sinh ^{2}x=1$.

Solution

$$\begin{eqnarray*} \cosh ^{2}x-\sinh ^{2}x&=&\left( \dfrac{e^{x}+e^{-x}}{2}\right) ^{2}-\left( \dfrac{e^{x}-e^{-x}}{2}\right) ^{2}\\ &=&\dfrac{e^{2x}+2e^{0}+e^{-2x}}{4}-\dfrac{ e^{2x}-2e^{0}+e^{-2x}}{4}=\dfrac{2+2}{4}=1 \end{eqnarray*}$$