Verify the identity.

$tan{h}^2(x)+sec{h}^2(x)=1$

First, use the definitions of the hyperbolic functions to write the expression in exponential terms, then simplify.

$tan{h}^2(x)+sec{h}^2(x)=\frac{(-e^{-x}{)}^2+(\frac{e^x+e^{-x}}{e^x+e^{-x}}{)}^2}{?}(e^x+e^{-x}{)}^2$