Question

how can it be cosh^2y - sinh^2y =1 ?

Answer 1

\[coshy = \frac{e^y+e^{-y}}{2}\]\[cosh^2y = (\frac{e^y+e^{-y}}{2})^2 = \frac{e^{2y}+e^{-2y}+2}{4}\]\[sinhy = \frac{e^y-e^{-y}}{2}\]\[sinh^2y = (\frac{e^{y}-e^{-y}}{2})^2 = \frac{e^{2y}+e^{-2y}-2}{4}\]\[cosh^2y-sinh^2y = \frac{e^{2y}+e^{-2y}+2}{4} - \frac{e^{2y}+e^{-2y}-2}{4}=\frac{4}{4}=1\]

Answer 2

thanks