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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

$(a)\;0\qquad(b)\;2\pi\qquad(c)\;\pi\qquad(d)\;4\pi$

1 Answer

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

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