If {tan ^{ - 1}}frac{{1 - x}}{{1 + x}} = frac | vedclass

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(C) Given: ${	an ^{ - 1}}\left( \frac{1 - x}{1 + x} ight) = \frac{1}{2}{	an ^{ - 1}}x$Let $x = 	an 	heta$,then $	heta = {	an ^{ - 1}}x$.Substi

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$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(C) Given: ${	an ^{ - 1}}\left( \frac{1 - x}{1 + x} ight) = \frac{1}{2}{	an ^{ - 1}}x$Let $x = 	an 	heta$,then $	heta = {	an ^{ - 1}}x$.Substituting $x = 	an 	heta$ in the equation:${	an ^{ - 1}}\left( \frac{1 - 	an 	heta}{1 + 	an 	heta} ight) = \frac{1}{2}	heta$Using the formula $	an \left( \frac{\pi}{4} - 	heta ight) = \frac{	an \frac{\pi}{4} - 	an 	heta}{1 + 	an \frac{\pi}{4}	an 	heta} = \frac{1 - 	an 	heta}{1 + 	an 	heta}$:${	an ^{ - 1}}\left( 	an \left( \frac{\pi}{4} - 	heta ight) ight) = \frac{	heta}{2}$$\frac{\pi}{4} - 	heta = \frac{	heta}{2}$$\frac{\pi}{4} = 	heta + \frac{	heta}{2} = \frac{3	heta}{2}$$	heta = \frac{\pi}{4} 	imes \frac{2}{3} = \frac{\pi}{6}$Since $x = 	an 	heta$,we have $x = 	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

If ${\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x$,then $x = $

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