MCQ
$2{\tan ^{ - 1}}(\cos x) = {\tan ^{ - 1}}({\rm{cose}}{{\rm{c}}^2}x),$ તો $ x =$
- A$\frac{\pi }{2}$
- B$\pi $
- C$\frac{\pi }{6}$
- ✓$\frac{\pi }{3}$
==> ${\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{1 - {{\cos }^2}x}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{{{{\sin }^2}x}}} \right)$
$ \Rightarrow \frac{{2\cos x}}{{{{\sin }^2}x}} = \frac{1}{{{{\sin }^2}x}}$
==> $2\cos x = 1$
$ \Rightarrow x = \frac{\pi }{3}$.
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