Answer
$0$
Work Step by Step
Use hyperbolic functions formula as follows:
$ \cosh x= \dfrac{e^x+e^{-x}}{2}$ and $ \sinh x= \dfrac{e^x-e^{-x}}{2}$
Need to solve: $\ln (\cosh x+\sinh x )+ \ln (\cosh x -\sinh x )$
This implies, $\ln (\cosh x+\sinh x )+ \ln (\cosh x -\sinh x )=\ln [\dfrac{e^x+e^{-x}}{2}+\dfrac{e^x-e^{-x}}{2}]+\ln [\dfrac{e^x+e^{-x}}{2} -\dfrac{e^x-e^{-x}}{2}]=\ln [\dfrac{2e^x}{2}] + \ln [\dfrac{2e^{-x}}{2}]=x-x$
Hence, $\ln (\cosh x+\sinh x )+ \ln (\cosh x -\sinh x )=0$
