Answer
$(\cosh x+\sinh x)^{n} =\cosh nx +\ sinh nx$
Work Step by Step
As we know, $\cosh(x) = \dfrac{e^{x} + e^{-x}}{2}$ and
$\sinh(x) = \dfrac{e^{x} - e^{-x}}{2}$
Thus, $(\dfrac{e^{x} + e^{-x}}{2} + \dfrac{e^{x} - e^{-x}}{2} )^{n}=(\dfrac{2e^{x}}{2})^{n}$
or, $= (e^{x})^{n}$
or, $=e^{nx}$
or, $=\dfrac{2e^{nx}}{2}$
Thus, $\dfrac{e^{nx} + e^{-nx}}{2}+\dfrac{e^{nx} - e^{-nx}}{2}=\cosh nx +\sinh nx$
