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If $\sin^{-1}\frac{2a}{1+a^2} +\cos^{-1}\frac{1-a^2}{1+a^2}=\tan^{-1}\frac{2x}{1-x^2},where\; a,x,\in\;[0,1],$then the value of x is

$(a)\;0\qquad(b)\;\large\frac{a}{2}\qquad(c)\;a\qquad(d)\;\large\frac{2a}{1-a^{2}}$

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