A) 1
B) \[\sqrt{3}\]
C) \[\frac{1}{\sqrt{3}}\]
D) None of these
Correct Answer: C
Solution :
We have \[{{\tan }^{-1}}\,\frac{1-x}{1+x}=\frac{1}{2}{{\tan }^{-1}}x\] \[\Rightarrow \,\,{{\tan }^{-1}}\,\left[ \frac{1-\tan \theta }{1+\tan \theta } \right]=\frac{1}{2}\theta \] (Putting \[x=\tan \theta )\] \[\Rightarrow \,\,{{\tan }^{-1}}\,\left[ \frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta } \right]=\frac{\theta }{2}\] \[\Rightarrow \,\,{{\tan }^{-1}}\tan \,\left( \frac{\pi }{4}-\theta \right)=\frac{\theta }{2}\,\,\Rightarrow \,\,\frac{\pi }{4}-\theta =\frac{\theta }{2}\] \[\Rightarrow \,\,\theta =\frac{\pi }{6}={{\tan }^{-1}}x\,\,\Rightarrow \,\,x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}\].
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