Answer
$[f(x)]^2-[g(x)]^2=\frac{4}{4}=1$
Work Step by Step
$[f(x)]^2-[g(x)]^2=(\frac{e^x+e^{-x}}{2})^2-(\frac{e^x-e^{-x}}{2})^2=\frac{(e^{x})^2+2e^xe^{-x}+(e^{-x})^2}{4}-\frac{(e^{x})^2-2e^xe^{-x}+(e^{-x})^2}{4}=\frac{e^{2x}+2+e^{-2x}}{4}-\frac{e^{2x}-2+e^{-2x}}{4}=\frac{4}{4}=1$
