If cot^{-1}[(cos alpha)^{1/2}] - tan^{-1}[(co | vedclass

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(A) Given: $\cot^{-1}[(\cos \alpha)^{1/2}] - 	an^{-1}[(\cos \alpha)^{1/2}] = x$Using the identity $\cot^{-1}(y) = 	an^{-1}(\frac{1}{y})$,we get:$	a

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

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(A) Given: $\cot^{-1}[(\cos \alpha)^{1/2}] - 	an^{-1}[(\cos \alpha)^{1/2}] = x$Using the identity $\cot^{-1}(y) = 	an^{-1}(\frac{1}{y})$,we get:$	an^{-1}[\frac{1}{\sqrt{\cos \alpha}}] - 	an^{-1}[\sqrt{\cos \alpha}] = x$Using the formula $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}(\frac{A-B}{1+AB})$:$	an^{-1}[\frac{\frac{1}{\sqrt{\cos \alpha}} - \sqrt{\cos \alpha}}{1 + (\frac{1}{\sqrt{\cos \alpha}})(\sqrt{\cos \alpha})}] = x$$	an^{-1}[\frac{\frac{1-\cos \alpha}{\sqrt{\cos \alpha}}}{1+1}] = x$$	an x = \frac{1-\cos \alpha}{2\sqrt{\cos \alpha}}$Now,we need to find $\sin x$. Using the identity $\sin x = \frac{	an x}{\sqrt{1+	an^2 x}}$ or by constructing a right triangle where the opposite side is $1-\cos \alpha$ and the adjacent side is $2\sqrt{\cos \alpha}$:Hypotenuse = $\sqrt{(1-\cos \alpha)^2 + (2\sqrt{\cos \alpha})^2} = \sqrt{1 - 2\cos \alpha + \cos^2 \alpha + 4\cos \alpha} = \sqrt{1 + 2\cos \alpha + \cos^2 \alpha} = \sqrt{(1+\cos \alpha)^2} = 1+\cos \alpha$Therefore,$\sin x = \frac{	ext{Opposite}}{	ext{Hypotenuse}} = \frac{1-\cos \alpha}{1+\cos \alpha} = \frac{2\sin^2(\alpha/2)}{2\cos^2(\alpha/2)} = 	an^2(\frac{\alpha}{2})$.

If $\cot^{-1}[(\cos \alpha)^{1/2}] - \tan^{-1}[(\cos \alpha)^{1/2}] = x$,then $\sin x = $

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