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If \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), then \(\log \left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)=\)
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Verified Answer
The correct answer is:
\(2 \tanh ^{-1}\left(\tan \frac{\theta}{2}\right)\)
\(\begin{array}{rlrl}
\text {Let } & \log \left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right) =x \\
\Rightarrow & \tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right) =e^x \\
\Rightarrow & \frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}} =e^x
\end{array}\)
On applying componendo and dividendo rule, we get
\(\begin{aligned}
2 & \frac{2 \tan \frac{\theta}{2}}{2} =\frac{e^x-1}{e^x+1} \\
\Rightarrow & \tan \frac{\theta}{2} =\frac{e^{x / 2}-e^{-x / 2}}{e^{x / 2}+e^{-x / 2}}=\tanh \left(\frac{x}{2}\right) \\
\Rightarrow & \frac{x}{2} =\tanh ^{-1}\left(\tan \frac{\theta}{2}\right) \\
\Rightarrow & x =2 \tanh ^{-1}\left(\tan \frac{\theta}{2}\right)
\end{aligned}\)
Hence, option (b) is correct.
\text {Let } & \log \left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right) =x \\
\Rightarrow & \tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right) =e^x \\
\Rightarrow & \frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}} =e^x
\end{array}\)
On applying componendo and dividendo rule, we get
\(\begin{aligned}
2 & \frac{2 \tan \frac{\theta}{2}}{2} =\frac{e^x-1}{e^x+1} \\
\Rightarrow & \tan \frac{\theta}{2} =\frac{e^{x / 2}-e^{-x / 2}}{e^{x / 2}+e^{-x / 2}}=\tanh \left(\frac{x}{2}\right) \\
\Rightarrow & \frac{x}{2} =\tanh ^{-1}\left(\tan \frac{\theta}{2}\right) \\
\Rightarrow & x =2 \tanh ^{-1}\left(\tan \frac{\theta}{2}\right)
\end{aligned}\)
Hence, option (b) is correct.
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