If 2 tan^{-1}(cos x) = tan^{-1}(csc^2 x),then | vedclass

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

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$\frac{\pi}{3}$

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(D) Given the equation: $2 	an^{-1}(\cos x) = 	an^{-1}(\csc^2 x)$.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}(\csc^2 x)$.Since $1 - \cos^2 x = \sin^2 x$,the equation becomes:$	an^{-1}\left(\frac{2 \cos x}{\sin^2 x}ight) = 	an^{-1}\left(\frac{1}{\sin^2 x}ight)$.Equating the arguments:$\frac{2 \cos x}{\sin^2 x} = \frac{1}{\sin^2 x}$.Since $\sin^2 x 
eq 0$ (as $\csc^2 x$ is defined),we can cancel $\sin^2 x$ from both sides:$2 \cos x = 1$.$\cos x = \frac{1}{2}$.Thus,$x = \frac{\pi}{3}$.

If $2 \tan^{-1}(\cos x) = \tan^{-1}(\csc^2 x)$,then $x =$

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