Find the value of tan frac{1}{2} left[ sin^{- | vedclass

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(D) Let $x = 	an \alpha$. Then $\alpha = 	an^{-1} x$.
$\sin^{-1} \left( \frac{2x}{1+x^2} ight) = \sin^{-1} \left( \frac{2 	an \alpha}{1+	an^2 \

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$\frac{x+y}{1-xy}$

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(D) Let $x = 	an \alpha$. Then $\alpha = 	an^{-1} x$.
$\sin^{-1} \left( \frac{2x}{1+x^2} ight) = \sin^{-1} \left( \frac{2 	an \alpha}{1+	an^2 \alpha} ight) = \sin^{-1} (\sin 2\alpha) = 2\alpha = 2 	an^{-1} x$.
Let $y = 	an \beta$. Then $\beta = 	an^{-1} y$.
$\cos^{-1} \left( \frac{1-y^2}{1+y^2} ight) = \cos^{-1} \left( \frac{1-	an^2 \beta}{1+	an^2 \beta} ight) = \cos^{-1} (\cos 2\beta) = 2\beta = 2 	an^{-1} y$.
Substituting these into the expression:
$	an \frac{1}{2} [2 	an^{-1} x + 2 	an^{-1} y] = 	an [	an^{-1} x + 	an^{-1} y]$.
Using the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x+y}{1-xy} ight)$:
$= 	an [	an^{-1} \left( \frac{x+y}{1-xy} ight)] = \frac{x+y}{1-xy}$.

Find the value of $\tan \frac{1}{2} \left[ \sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2} \right]$,where $|x| <1, y>0$ and $xy < 1$.

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