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

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$\sqrt{2}$

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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 $1 - \cos^2 x = \sin^2 x$, the equation becomes:$\frac{2 \cos x}{\sin^2 x} = \frac{2}{\sin x}$.Assuming $\sin x 
eq 0$, we can simplify:$\frac{\cos x}{\sin x} = 1 \Rightarrow \cot x = 1$.Thus, $x = \frac{\pi}{4}$.Now, calculate $\sin x + \cos x$:$\sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

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

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