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

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

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$\frac{\pi}{4}$

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(B) Given the equation: $2	an^{-1}(\cos x) = 	an^{-1}(2\csc x)$.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}(2\csc x)$.Since $1-\cos^2 x = \sin^2 x$,this simplifies to:$\frac{2\cos x}{\sin^2 x} = 2\csc x$.Substituting $\csc x = \frac{1}{\sin x}$:$\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}$.Assuming $\sin x 
eq 0$,we multiply both sides by $\sin^2 x$:$2\cos x = 2\sin x$.Dividing by $2\cos x$ (assuming $\cos x 
eq 0$):$	an x = 1$.Thus,$x = \frac{\pi}{4}$.

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

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