MCQ
If $2{\tan ^{ - 1}}(\cos x) = {\tan ^{ - 1}}(2{\rm{cosec }}x),$ then $x =$
- A$\frac{{3\pi }}{4}$
- ✓$\frac{\pi }{4}$
- C$\frac{\pi }{3}$
- DNone of these
==> ${\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{1 - {{\cos }^2}x}}} \right)$$= {\tan ^{ - 1}}(2\,{\rm{cosec }}x)$
$\frac{{2\cos x}}{{{{\sin }^2}x}} = 2{\rm{cosec}}\,x \Rightarrow 2\cos x = 2\sin x$
or $\sin x = \cos x$
$ \Rightarrow x = \frac{\pi }{4}$.
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