If 2 tan^{-1}(cos x) = tan^{-1}(2 operatornam | vedclass

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(D) Given the equation: $2 	an^{-1}(\cos x) = 	an^{-1}(2 \operatorname{cosec} x)$.Using the formula $2 	an^{-1}(	heta) = 	an^{-1}\left(\frac{2	h

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$\frac{\pi}{4}$

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(D) Given the equation: $2 	an^{-1}(\cos x) = 	an^{-1}(2 \operatorname{cosec} x)$.Using the formula $2 	an^{-1}(	heta) = 	an^{-1}\left(\frac{2	heta}{1-	heta^2}ight)$,we get:$	an^{-1}\left(\frac{2 \cos x}{1 - \cos^2 x}ight) = 	an^{-1}\left(\frac{2}{\sin x}ight)$.Since $1 - \cos^2 x = \sin^2 x$,the equation becomes:$\frac{2 \cos x}{\sin^2 x} = \frac{2}{\sin x}$.Assuming $\sin x 
eq 0$,we can simplify:$\frac{\cos x}{\sin x} = 1$,which implies $\cot x = 1$.Thus,$x = \frac{\pi}{4}$.Checking the value: $2 	an^{-1}(\cos(\frac{\pi}{4})) = 2 	an^{-1}(\frac{1}{\sqrt{2}})$ and $	an^{-1}(2 \operatorname{cosec}(\frac{\pi}{4})) = 	an^{-1}(2 \sqrt{2})$.Since $2 	an^{-1}(\frac{1}{\sqrt{2}}) = 	an^{-1}(\frac{2(1/\sqrt{2})}{1-1/2}) = 	an^{-1}(\frac{\sqrt{2}}{1/2}) = 	an^{-1}(2\sqrt{2})$,the solution is correct.

If $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \operatorname{cosec} x)$,then the value of $x$ is

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