Answer
\[\frac{1+\tanh x}{1-\tanh x}=e^{2x}\]
Work Step by Step
To show that :- \[\frac{1+\tanh x}{1-\tanh x}=e^{2x}\]
We know that \[\sinh x=\frac{e^x-e^{-x}}{2}\;\;\;...(1)\]
and
\[\cosh x=\frac{e^x+e^{-x}}{2}\;\;\;...(2)\]
\[\tanh x=\frac{\sinh x}{\cosh x}\]
Using (1) and (2)
\[\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}\;\;\;...(3)\]
Consider \[\frac{1+\tanh x}{1-\tanh x}\]
Using (3)
\[\frac{1+\tanh x}{1-\tanh x}=\frac{1+\frac{e^x-e^{-x}}{e^x+e^{-x}}}{1-\frac{e^x-e^{-x}}{e^x+e^{-x}}}\]
\[\Rightarrow \frac{1+\tanh x}{1-\tanh x}=\frac{(e^x+e^{-x})+(e^x-e^{-x})}{(e^x+e^{-x})-(e^x-e^{-x})}\]
\[\Rightarrow \frac{1+\tanh x}{1-\tanh x}=\frac{2e^x}{2e^{-x}}\]
\[\Rightarrow \frac{1+\tanh x}{1-\tanh x}=e^{2x}\]
Hence, \[\frac{1+\tanh x}{1-\tanh x}=e^{2x}\]
