frac{1+tanh left(frac{x}{2}right)}{1-tanh lef | vedclass

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(B) We know that $	anh \left(\frac{x}{2}ight) = \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}$.Substituting this into the expression:$\frac{1 + 	a

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$e^x$

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(B) We know that $	anh \left(\frac{x}{2}ight) = \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}$.Substituting this into the expression:$\frac{1 + 	anh \left(\frac{x}{2}ight)}{1 - 	anh \left(\frac{x}{2}ight)} = \frac{1 + \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}}{1 - \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}}$$= \frac{\frac{e^{x/2} + e^{-x/2} + e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}}{\frac{e^{x/2} + e^{-x/2} - (e^{x/2} - e^{-x/2})}{e^{x/2} + e^{-x/2}}}$$= \frac{2e^{x/2}}{2e^{-x/2}}$$= e^{x/2 - (-x/2)} = e^x$.

$\frac{1+\tanh \left(\frac{x}{2}\right)}{1-\tanh \left(\frac{x}{2}\right)} = $

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