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If $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x$, then $x$ has the value
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$\frac{1}{\sqrt{3}}$
$\begin{aligned} & \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x \\ & \Rightarrow \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\frac{1}{2} \tan ^{-1}(\tan \theta) \\ & \Rightarrow \tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\theta\right)\right)=\frac{1}{2} \theta \\ & \Rightarrow \frac{\pi}{4}-\theta=\frac{1}{2} \theta \\ & \Rightarrow \frac{\pi}{4}=\frac{3 \theta}{2} \\ & \Rightarrow \theta=\frac{\pi}{6} \\ & \Rightarrow x=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\end{aligned}$
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