Using the definitions of the corresponding functions, prove…

Using the definitions of the corresponding functions, prove: $\ \cosh^2z-\sinh^2z=1$.

Step 1 · Square each function

$$ \cosh^2z=\frac{e^{2z}+2+e^{-2z}}4, \qquad \sinh^2z=\frac{e^{2z}-2+e^{-2z}}4 $$

Step 2 · Subtract

$$ \cosh^2z-\sinh^2z=\frac{(e^{2z}+2+e^{-2z})-(e^{2z}-2+e^{-2z})}4=\frac44=1 $$

Alternative proof · factor as a difference of squares

$$ \cosh z+\sinh z=\frac{2e^z}2=e^z, \qquad \cosh z-\sinh z=\frac{2e^{-z}}2=e^{-z} $$

$$ (\cosh z+\sinh z)(\cosh z-\sinh z)=\cosh^2z-\sinh^2z=e^z\cdot e^{-z}=e^0=1 $$

Connection to the circular identity

Using $\cosh(iz)=\cos z,\ \sinh(iz)=i\sin z$: substituting $z\to iz$ recovers $\cos^2z-(i\sin z)^2=\cos^2z+\sin^2z=1$ — the hyperbolic and circular Pythagorean identities are the same statement viewed through a rotation by $i$.