Question

Prove that ${\cosh }^{2}x-{\sinh }^{2}x=1$

Answer 1

Given,

$cos{h}^{2}x-sin{h}^{2}x$

$={\left(\frac{e^x+e^{-x}}{2}\right)}^2$- ${\left(\frac{e^x+e^{-x}}{2}\right)}^2$

$=\frac{e^{2x+e^{-2x+2}}}{4}-\frac{e^{2x+e^{-2x+2}}}{4}$

$=\frac{e^{2x}+e^{-2x}+2-e^{2x}-e^{2x}+2}{4}$

$=\frac{4}{4}$

$=1$