MCQ
If $2tan^{-1}(cosx) = tan^{-1}(cosec^2x)$ then $x =$
- A$\frac{\pi}{2}$
- B$\pi$
- C$\frac{\pi}{6}$
- ✓$\frac{\pi}{3}$
$\Rightarrow \frac{2 \cos x}{\sin ^{2} x}=\frac{1}{\sin ^{2} x} \Rightarrow 2 \cos x=1 \Rightarrow x=\frac{\pi}{3}$
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