MCQ
If ${\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x$, then $ x =$
- A$1$
- B$\sqrt 3 $
- ✓$\frac{1}{{\sqrt 3 }}$
- DNone of these
$ \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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