frac{1+tanh frac{x}{2}}{1-tanh frac{x}{2}} is | vedclass

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(B) We know that $	anh 	heta = \frac{\sinh 	heta}{\cosh 	heta} = \frac{e^{	heta} - e^{-	heta}}{e^{	heta} + e^{-	heta}}$.Substituting $	heta =

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$e^{x}$

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(B) We know that $	anh 	heta = \frac{\sinh 	heta}{\cosh 	heta} = \frac{e^{	heta} - e^{-	heta}}{e^{	heta} + e^{-	heta}}$.Substituting $	heta = \frac{x}{2}$,we have $	anh \frac{x}{2} = \frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}}$.Now,consider the expression $\frac{1+	anh \frac{x}{2}}{1-	anh \frac{x}{2}}$:$= \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}}$$= \frac{e^{x/2}}{e^{-x/2}} = e^{x/2 + x/2} = e^x$.

$\frac{1+\tanh \frac{x}{2}}{1-\tanh \frac{x}{2}}$ is equal to

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