Pythagorean identity for sinh and cosh

Theorem

The following formula holds: $$\cosh^2(z)-\sinh^2(z)=1,$$ where $\cosh$ denotes the hyperbolic cosine and $\sinh$ denotes the hyperbolic sine.

Proof

From the definitions $$\cosh(z)=\dfrac{e^{z}+e^{-z}}{2}$$ and $$\sinh(z)=\dfrac{e^{z}-e^{-z}}{2},$$ we see $$\begin{array}{ll} \cosh^2(z) - \sinh^2(z) &= \left( \dfrac{e^{z}+e^{-z}}{2} \right)^2 - \left( \dfrac{e^{z}-e^{-z}}{2} \right)^2 \\ &= \dfrac{1}{4} \left( e^{2z}+2+e^{-2z}-e^{2z}+2-e^{-2z} \right) \\ &= 1, \end{array}$$ as was to be shown. █