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$\frac{1+\tanh \frac{x}{2}}{1-\tanh \frac{x}{2}}$ is equal to
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The correct answer is:
$e^{x}$
$\begin{aligned} \frac{1+\tanh \frac{x}{2}}{1-\tanh \frac{x}{2}} & =\frac{\cosh \frac{x}{2}+\sinh \frac{x}{2}}{\cosh \frac{x}{2}-\sinh \frac{x}{2}} \\ & =\frac{e^{x / 2}}{e^{-x / 2}}=e^x\end{aligned}$
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