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If $y=\log _e \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)$, then $\tan h\left(\frac{y}{2}\right)=$
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Verified Answer
The correct answer is:
$\tan \frac{x}{2}$
It is given that $y=\log _c \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)$
$\Rightarrow \quad e^y=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}$
$\tan h\left(\frac{y}{2}\right)=\frac{e^{\frac{y}{2}}-e^{-\frac{y}{2}}}{e^{\frac{y}{2}}+e^{-\frac{y}{2}}}=\frac{e^y-1}{e^y+1}$
$\begin{aligned} & =\frac{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}-1}{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}+1}=\frac{1+\tan \frac{x}{2}-1+\tan \frac{x}{2}}{1+\tan \frac{x}{2}+1-\tan \frac{x}{2}} \\ & =\frac{2 \tan \frac{x}{2}}{2}=\tan \frac{x}{2} \\ & \end{aligned}$
Hence, option (d) is correct.
$\Rightarrow \quad e^y=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}$
$\tan h\left(\frac{y}{2}\right)=\frac{e^{\frac{y}{2}}-e^{-\frac{y}{2}}}{e^{\frac{y}{2}}+e^{-\frac{y}{2}}}=\frac{e^y-1}{e^y+1}$
$\begin{aligned} & =\frac{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}-1}{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}+1}=\frac{1+\tan \frac{x}{2}-1+\tan \frac{x}{2}}{1+\tan \frac{x}{2}+1-\tan \frac{x}{2}} \\ & =\frac{2 \tan \frac{x}{2}}{2}=\tan \frac{x}{2} \\ & \end{aligned}$
Hence, option (d) is correct.
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