If tan ^{-1}left(frac{1-x}{1+x}right)=frac{1} | vedclass

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(D) Given equation: $	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 $

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

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(D) Given equation: $	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(\frac{\pi}{4} - 	heta) = \frac{1-	an 	heta}{1+	an 	heta}$:$	an ^{-1}\left(	an \left(\frac{\pi}{4}-	hetaight)ight)=\frac{1}{2} 	heta$$\frac{\pi}{4}-	heta=\frac{1}{2} 	heta$$\frac{\pi}{4}=\frac{3 	heta}{2}$$	heta=\frac{\pi}{6}$Since $x = 	an 	heta$,we have $x = 	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

If $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x$,then $x$ has the value

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