Question

If $2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left({\text{cosec}}^{2}x\right),$ where x $ϵ(0,\pi )$, then $x$ is equal to

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (D)

$\frac{\pi }{3}$

Given, $2{\tan }^{-1}\left(\cos x\right)={\tan }^{-1}\left({cosec}^{2}x\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{2\cos x}{1-{\cos }^{2}x}\right)={\tan }^{-1}\left({cosec}^{2}x\right)$
$\{\therefore -1<\cos x<1\}$
$\Rightarrow \frac{2\cos x}{{\sin }^{2}x}=\frac{1}{{\sin }^{2}x}\Rightarrow 2\cos x=1$

$\Rightarrow x=\frac{\pi }{3}$