Answer
1
Work Step by Step
We apply hyperbolic function formulas:
$ \cosh x= \dfrac{e^x+e^{-x}}{2}$ and $ \sinh x= \dfrac{e^x-e^{-x}}{2}$
Thus, $\cosh^2 x-\sinh^2x=(\cosh x -\sinh x) (\cosh x +\sinh x)=[(\dfrac{e^x+e^{-x}}{2})-(\dfrac{e^x-e^{-x}}{2})][(\dfrac{e^x+e^{-x}}{2})-(\dfrac{e^x-e^{-x}}{2})]=e^x \cdot e^{-x}=e^{0}=1$
