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

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Aicharya · Accepted Answer

Given, $y=\log _e \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)$ $\begin{aligned} & =\log _e\left(\frac{\tan \left(\frac{\pi}{4}\right)+\tan \frac{x}{2}}{1-\tan \left(\frac{\pi}{4}\right) \tan \left(\frac{x}{2}\right)}\right)=\log _e\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right) \ \Rightarrow \quad & e^y=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}} \Rightarrow \frac{e^y+1}{e^y-1}=\frac{2}{2 \tan \frac{x}{2}} \ \Rightarrow \quad & \quad \frac{e^y-1}{e^y+1}=\tan \frac{x}{2} \Rightarrow \tan h\left(\frac{y}{2}\right)=\tan \left(\frac{x}{2}\right) \end{aligned}$

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

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