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

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(D) Given,$y = \log_e 	an \left(\frac{\pi}{4} + \frac{x}{2}ight)$
Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we hav

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$	an \left(\frac{x}{2}ight)$

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(D) Given,$y = \log_e 	an \left(\frac{\pi}{4} + \frac{x}{2}ight)$
Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:
$ \quad=\log _e\left(\frac{	an \left(\frac{\pi}{4}ight)+	an \frac{x}{2}}{1-	an \left(\frac{\pi}{4}ight) 	an \left(\frac{x}{2}ight)}ight)=\log _e\left(\frac{1+	an \frac{x}{2}}{1-	an \frac{x}{2}}ight)$
Taking the exponential on both sides:
$e^y = \frac{1 + 	an(x/2)}{1 - 	an(x/2)}$
Using the componendo and dividendo rule:
$\frac{e^y - 1}{e^y + 1} = \frac{(1 + 	an(x/2)) - (1 - 	an(x/2))}{(1 + 	an(x/2)) + (1 - 	an(x/2))} = \frac{2 	an(x/2)}{2} = 	an \left(\frac{x}{2}ight)$
Since $	anh \left(\frac{y}{2}ight) = \frac{e^y - 1}{e^y + 1}$,we get:
$	anh \left(\frac{y}{2}ight) = 	an \left(\frac{x}{2}ight)$

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

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