A) \[\frac{1}{2}\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[\sqrt{3}\]
D) 2
Correct Answer: B
Solution :
\[\because \] \[{{\tan }^{-1}}\frac{1-x}{1+x}=\frac{1}{2}{{\tan }^{-1}}x\] Let \[x=\tan \theta \] \[\therefore \] \[{{\tan }^{-1}}\left( \frac{1-\tan \theta }{1+\tan \theta } \right)=\frac{1}{2}{{\tan }^{-1}}(tan\theta )\] \[\Rightarrow \]\[{{\tan }^{-1}}\left( \tan \left( \frac{\pi }{4}-\theta \right) \right)=\frac{1}{2}{{\tan }^{-1}}(tan\theta )\] \[\Rightarrow \] \[\frac{\pi }{4}-\theta =\frac{\theta }{2}\] \[\Rightarrow \] \[\frac{3\theta }{2}=\frac{\pi }{4}\Rightarrow \theta =\frac{\pi }{6}.\] \[\therefore \] \[x=\tan \theta =\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}\]
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