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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:
$\frac{\pi^c}{4}$
We have $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$
$$
\begin{aligned}
& \therefore \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}\left(\frac{2}{\sin x}\right) \\
& \therefore \frac{2 \cos x}{\sin ^2 x}=\frac{2}{\sin x} \Rightarrow \sin x \cos x=\sin ^2 x \\
& \therefore \sin x(\sin x-\cos x)=0 \Rightarrow \sin x=0 \text { or } \tan x=1 \\
& \therefore x=0 \text { or } x=\frac{\pi}{4}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}\left(\frac{2}{\sin x}\right) \\
& \therefore \frac{2 \cos x}{\sin ^2 x}=\frac{2}{\sin x} \Rightarrow \sin x \cos x=\sin ^2 x \\
& \therefore \sin x(\sin x-\cos x)=0 \Rightarrow \sin x=0 \text { or } \tan x=1 \\
& \therefore x=0 \text { or } x=\frac{\pi}{4}
\end{aligned}
$$
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