lean_workbook_plus_16252

Using the definitions of $\sinh x$ and $\cosh x$ in terms of $e^{x}$ and $e^{-x}$ , show that\n\n(a) $\cosh(\ln a) = \frac{a^{2} + 1}{2a}$ , where $a>0$ ,\n\n(b) $\cosh x \cosh y - \sinh x \sinh y \equiv \cosh (x-y)$ .