MCQ
If ${\tan ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3},$ then $x =$
- A$\sqrt 2 $
- B$3$
- ✓$\sqrt 3 $
- D$\frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$
${\tan ^{ - 1}}x + {\cot ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{{2\pi }}{3}$
==> ${\cot ^{ - 1}}x = \frac{{2\pi }}{3} - \frac{\pi }{2}$ = $\frac{\pi }{6}$
==> $x = \sqrt 3 $.
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