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}$.Substituting $	heta = \frac{x}{2}$, we have $	anh \frac{x}{2} = \frac{\sinh \fra

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$e^{x}$

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(B) We know that $	anh 	heta = \frac{\sinh 	heta}{\cosh 	heta}$.Substituting $	heta = \frac{x}{2}$, we have $	anh \frac{x}{2} = \frac{\sinh \frac{x}{2}}{\cosh \frac{x}{2}}$.Substituting this into the expression:$\frac{1+	anh \frac{x}{2}}{1-	anh \frac{x}{2}} = \frac{1+\frac{\sinh \frac{x}{2}}{\cosh \frac{x}{2}}}{1-\frac{\sinh \frac{x}{2}}{\cosh \frac{x}{2}}}$$= \frac{\cosh \frac{x}{2} + \sinh \frac{x}{2}}{\cosh \frac{x}{2} - \sinh \frac{x}{2}}$Using the definitions $\cosh 	heta = \frac{e^{	heta} + e^{-	heta}}{2}$ and $\sinh 	heta = \frac{e^{	heta} - e^{-	heta}}{2}$, we get:$\cosh \frac{x}{2} + \sinh \frac{x}{2} = e^{x/2}$$\cosh \frac{x}{2} - \sinh \frac{x}{2} = e^{-x/2}$Therefore, the expression becomes $\frac{e^{x/2}}{e^{-x/2}} = e^{x/2 - (-x/2)} = e^x$.

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

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