If tan (cot x) = cot (tan x), then sin 2x = | vedclass

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(B) Given the equation: $	an (\cot x) = \cot (	an x)$Using the identity $\cot 	heta = 	an (\frac{\pi}{2} - 	heta)$,we have:$	an (\cot x) = 	an

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$\frac{4}{(2n + 1)\pi}$

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(B) Given the equation: $	an (\cot x) = \cot (	an x)$Using the identity $\cot 	heta = 	an (\frac{\pi}{2} - 	heta)$,we have:$	an (\cot x) = 	an (\frac{\pi}{2} - 	an x)$The general solution for $	an \alpha = 	an \beta$ is $\alpha = n\pi + \beta$,where $n \in \mathbb{Z}$.Thus,$\cot x = n\pi + \frac{\pi}{2} - 	an x$Rearranging the terms:$\cot x + 	an x = n\pi + \frac{\pi}{2}$Using the identity $	an x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\frac{1}{2} \sin 2x} = \frac{2}{\sin 2x}$:$\frac{2}{\sin 2x} = n\pi + \frac{\pi}{2} = \frac{(2n + 1)\pi}{2}$Solving for $\sin 2x$:$\sin 2x = \frac{4}{(2n + 1)\pi}$

If $\tan (\cot x) = \cot (\tan x),$ then $\sin 2x =$

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