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

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(C) दिया गया है: ${	an ^{ - 1}}\left( \frac{1 - x}{1 + x} ight) = \frac{1}{2}{	an ^{ - 1}}x$माना $x = 	an 	heta$,तो $	heta = {	an ^{ - 1}}x$.स

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) दिया गया है: ${	an ^{ - 1}}\left( \frac{1 - x}{1 + x} ight) = \frac{1}{2}{	an ^{ - 1}}x$माना $x = 	an 	heta$,तो $	heta = {	an ^{ - 1}}x$.समीकरण में $x = 	an 	heta$ रखने पर:${	an ^{ - 1}}\left( \frac{1 - 	an 	heta}{1 + 	an 	heta} ight) = \frac{1}{2}	heta$सूत्र $	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}$ का उपयोग करने पर:${	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}$चूंकि $x = 	an 	heta$,इसलिए $x = 	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

यदि ${\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x$ है,तो $x = $

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