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If $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$, then the value of $x$ is
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Verified Answer
The correct answer is:
$\frac{\pi}{4}$
$$
\begin{aligned}
& 2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x) \\
\Rightarrow & \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)=\tan ^{-1}(2 \operatorname{cosec} x) \\
\Rightarrow & \frac{2 \cos x}{1-\cos ^{2} x}=2 \operatorname{cosec} x \\
\Rightarrow & \frac{2 \cos x}{\sin ^{2} x}=2 \operatorname{cosec} x \\
\Rightarrow & \sin x=\cos x \Rightarrow x=\frac{\pi}{4}
\end{aligned}
$$
\begin{aligned}
& 2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x) \\
\Rightarrow & \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)=\tan ^{-1}(2 \operatorname{cosec} x) \\
\Rightarrow & \frac{2 \cos x}{1-\cos ^{2} x}=2 \operatorname{cosec} x \\
\Rightarrow & \frac{2 \cos x}{\sin ^{2} x}=2 \operatorname{cosec} x \\
\Rightarrow & \sin x=\cos x \Rightarrow x=\frac{\pi}{4}
\end{aligned}
$$
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