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If $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x$, then $x$ is
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The correct answer is:
$\frac{1}{\sqrt{3}}$
$\begin{aligned} & \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x \\ & \Rightarrow \tan ^{-1}(1)-\tan ^{-1}(x)=\frac{1}{2} \tan ^{-1} x \\ & \Rightarrow \frac{\pi}{4}=\frac{3}{2} \tan ^{-1} x \\ & \Rightarrow x=\tan \left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}\end{aligned}$
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