MCQ
If ${\tan ^{ - 1}}\frac{{a + x}}{a} + {\tan ^{ - 1}}\frac{{a - x}}{a} = \frac{\pi }{6}$,then ${x^2} =$
- A$2\sqrt 3 a$
- B$\sqrt 3 a$
- ✓$2\sqrt 3 {a^2}$
- DNone of these
$ \Rightarrow \,{\tan ^{ - 1\,}}\left( {\frac{{\frac{{a + x}}{a} + \frac{{a - x}}{a}}}{{1 - \frac{{a + x}}{a}.\,\frac{{a - x}}{a}}}} \right) = \frac{\pi }{6}$
$ \Rightarrow \,\,\frac{{2{a^2}}}{{{x^2}}} = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}\,\, $
$\Rightarrow \,\,{x^2} = 2\sqrt 3 {a^2}$.
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