If cot ^{-1}(sqrt{cos alpha})-tan ^{-1}(sqrt{ | vedclass

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(D) Given the equation: $\cot ^{-1}(\sqrt{\cos \alpha})-	an ^{-1}(\sqrt{\cos \alpha})=x$. We know the identity $\cot ^{-1}(y) = \frac{\pi}{2} - 	an

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$	an ^2 \frac{\alpha}{2}$

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(D) Given the equation: $\cot ^{-1}(\sqrt{\cos \alpha})-	an ^{-1}(\sqrt{\cos \alpha})=x$. We know the identity $\cot ^{-1}(y) = \frac{\pi}{2} - 	an ^{-1}(y)$. Substituting this into the equation: $(\frac{\pi}{2} - 	an ^{-1}(\sqrt{\cos \alpha})) - 	an ^{-1}(\sqrt{\cos \alpha}) = x$. This simplifies to: $\frac{\pi}{2} - 2 	an ^{-1}(\sqrt{\cos \alpha}) = x$. Rearranging gives: $2 	an ^{-1}(\sqrt{\cos \alpha}) = \frac{\pi}{2} - x$. Taking the tangent of both sides: $	an(2 	an ^{-1}(\sqrt{\cos \alpha})) = 	an(\frac{\pi}{2} - x) = \cot x$. Using the formula $	an(2	heta) = \frac{2 	an 	heta}{1 - 	an^2 	heta}$,where $	heta = 	an ^{-1}(\sqrt{\cos \alpha})$: $\cot x = \frac{2 \sqrt{\cos \alpha}}{1 - \cos \alpha}$. Using the half-angle identity $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$: $\cot x = \frac{2 \sqrt{\cos \alpha}}{2 \sin^2 \frac{\alpha}{2}} = \frac{\sqrt{\cos \alpha}}{\sin^2 \frac{\alpha}{2}}$. However,checking the standard identity $2 	an ^{-1}(y) = \cos ^{-1}(\frac{1-y^2}{1+y^2})$,we have $x = \frac{\pi}{2} - \cos ^{-1}(\frac{1-\cos \alpha}{1+\cos \alpha}) = \sin ^{-1}(\frac{1-\cos \alpha}{1+\cos \alpha})$. Thus,$\sin x = \frac{1-\cos \alpha}{1+\cos \alpha} = \frac{2 \sin^2 \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2}} = 	an^2 \frac{\alpha}{2}$.

If $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$,then the value of $\sin x$ is

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