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If $\tan ^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2} \tan ^{-1} x=0$, for $x>0$, then $x=$
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The correct answer is:
$\frac{1}{\sqrt{3}}$
Here $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x \Rightarrow \tan ^{-1}\left[\frac{1-x}{1+(1)(x)}\right]=\frac{1}{2} \tan ^{-1} x$
$\therefore \tan ^{-1}(1)-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x$
$\frac{\pi}{4}=\frac{3}{2} \tan ^{-1} x \Rightarrow \tan ^{-1} x=\frac{\pi}{4} \times \frac{2}{3} \Rightarrow \tan ^{-1} x=\frac{\pi}{6}$
$\therefore x=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$
$\therefore \tan ^{-1}(1)-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x$
$\frac{\pi}{4}=\frac{3}{2} \tan ^{-1} x \Rightarrow \tan ^{-1} x=\frac{\pi}{4} \times \frac{2}{3} \Rightarrow \tan ^{-1} x=\frac{\pi}{6}$
$\therefore x=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$
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