Answer
$\frac{1+tanh~x}{1-tanh~x} = e^{2x}$
Work Step by Step
$\frac{1+tanh~x}{1-tanh~x} = \frac{1+\frac{sinh~x}{cosh~x}}{1-\frac{sinh~x}{cosh~x}}$
$\frac{1+tanh~x}{1-tanh~x} = \frac{1+\frac{(e^x-e^{-x})/2}{(e^x+e^{-x})/2}}{1-\frac{(e^x-e^{-x})/2}{(e^x+e^{-x})/2}}$
$\frac{1+tanh~x}{1-tanh~x} = \frac{1+\frac{e^x-e^{-x}}{e^x+e^{-x}}}{1-\frac{e^x-e^{-x}}{e^x+e^{-x}}}$
$\frac{1+tanh~x}{1-tanh~x} = \frac{\frac{e^x+e^{-x}}{e^x+e^{-x}}+\frac{e^x-e^{-x}}{e^x+e^{-x}}}{\frac{e^x+e^{-x}}{e^x+e^{-x}}-\frac{e^x-e^{-x}}{e^x+e^{-x}}}$
$\frac{1+tanh~x}{1-tanh~x} = \frac{\frac{2e^x}{e^x+e^{-x}}}{\frac{2e^{-x}}{e^x+e^{-x}}}$
$\frac{1+tanh~x}{1-tanh~x} = \frac{2e^x}{2e^{-x}}$
$\frac{1+tanh~x}{1-tanh~x} = e^{2x}$
