If sin ^{-1}left(frac{2 a}{1+a^2}right)+cos ^ | vedclass

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(C) Given the equation: $\sin ^{-1}\left(\frac{2 a}{1+a^2}ight)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}ight)=	an ^{-1}\left(\frac{2 x}{1-x^2}ight)$

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$\frac{2 a}{1-a^2}$

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(C) Given the equation: $\sin ^{-1}\left(\frac{2 a}{1+a^2}ight)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}ight)=	an ^{-1}\left(\frac{2 x}{1-x^2}ight)$.We know the standard identities for $a \in (0, 1)$:$\sin ^{-1}\left(\frac{2 a}{1+a^2}ight) = 2 	an ^{-1} a$$\cos ^{-1}\left(\frac{1-a^2}{1+a^2}ight) = 2 	an ^{-1} a$Substituting these into the given equation:$2 	an ^{-1} a + 2 	an ^{-1} a = 	an ^{-1}\left(\frac{2 x}{1-x^2}ight)$$4 	an ^{-1} a = 2 	an ^{-1} x$$2 	an ^{-1} a = 	an ^{-1} x$Using the formula $2 	an ^{-1} a = 	an ^{-1}\left(\frac{2 a}{1-a^2}ight)$:$	an ^{-1}\left(\frac{2 a}{1-a^2}ight) = 	an ^{-1} x$Therefore,$x = \frac{2 a}{1-a^2}$.

If $\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ where $a, x \in(0,1)$,then the value of $x$ is

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