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If $x_1=2\tan^{-1}\big(\large\frac{1+x}{1-x}\big)$$,x_2=\sin^{-1}\big[\large\frac{1-x^2}{1+x^2}\big]$ where $x\in(0,1)$ then $x_1+x_2$ is equal to


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Here $x_1=2\tan^{-1}\big(\large\frac{1+x}{1-x}\big)$
$\qquad x_2=\sin^{-1}\big(\large\frac{1-x^2}{1+x^2}\big)$
Hence (c) is the correct answer.
answered Oct 11, 2013 by sreemathi.v
edited Jul 18, 2014 by sharmaaparna1

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