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

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(B) Given the equation: $\cot ^{-1}(\sqrt{\cos \alpha})-	an ^{-1}(\sqrt{\cos \alpha})=x$ Using the identity $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$

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$	an ^2\left(\frac{\alpha}{2}ight)$

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(B) Given the equation: $\cot ^{-1}(\sqrt{\cos \alpha})-	an ^{-1}(\sqrt{\cos \alpha})=x$ Using the identity $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$,we have: $	an ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}ight)-	an ^{-1}(\sqrt{\cos \alpha})=x$ Applying the formula $	an ^{-1}(A) - 	an ^{-1}(B) = 	an ^{-1}\left(\frac{A-B}{1+AB}ight)$: $	an ^{-1}\left[\frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}ight]=x$ $	an ^{-1}\left[\frac{\frac{1-\cos \alpha}{\sqrt{\cos \alpha}}}{1+1}ight]=x$ $	an ^{-1}\left(\frac{1-\cos \alpha}{2\sqrt{\cos \alpha}}ight)=x$ Thus,$	an x = \frac{1-\cos \alpha}{2\sqrt{\cos \alpha}}$. To find $\sin x$,we use the identity $\sin x = \frac{	an x}{\sqrt{1+	an^2 x}}$ or geometric construction. Alternatively,using the identity $\sin x = \frac{2	an(x/2)}{1+	an^2(x/2)}$ is complex here,so we use $\sin x = \frac{1-\cos \alpha}{1+\cos \alpha}$ directly from the simplification of the expression. $\sin x = \frac{2\sin^2(\alpha/2)}{2\cos^2(\alpha/2)} = 	an^2\left(\frac{\alpha}{2}ight)$.

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

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