If y = log_e tan left(frac{pi}{4} + frac{x}{2 | vedclass

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(D) Given $y = \log_e 	an \left(\frac{\pi}{4} + \frac{x}{2}ight)$.$e^y = 	an \left(\frac{\pi}{4} + \frac{x}{2}ight) = \frac{1 + 	an \frac{x}{2}

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$	an \frac{x}{2}$

Vedclass · Answer

(D) Given $y = \log_e 	an \left(\frac{\pi}{4} + \frac{x}{2}ight)$.$e^y = 	an \left(\frac{\pi}{4} + \frac{x}{2}ight) = \frac{1 + 	an \frac{x}{2}}{1 - 	an \frac{x}{2}}$.We know that $	anh \left(\frac{y}{2}ight) = \frac{e^y - 1}{e^y + 1}$.Substituting the value of $e^y$:$	anh \left(\frac{y}{2}ight) = \frac{\frac{1 + 	an \frac{x}{2}}{1 - 	an \frac{x}{2}} - 1}{\frac{1 + 	an \frac{x}{2}}{1 - 	an \frac{x}{2}} + 1} = \frac{(1 + 	an \frac{x}{2}) - (1 - 	an \frac{x}{2})}{(1 + 	an \frac{x}{2}) + (1 - 	an \frac{x}{2})} = \frac{2 	an \frac{x}{2}}{2} = 	an \frac{x}{2}$.Thus,option $D$ is correct.

If $y = \log_e \tan \left(\frac{\pi}{4} + \frac{x}{2}\right)$,then $\tanh \left(\frac{y}{2}\right) = $

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