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

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(C) Given equation is $	an ^{-1}\left(\frac{1-x}{1+x}ight)-\frac{1}{2} 	an ^{-1} x=0$. Rearranging the terms,we get $	an ^{-1}\left(\frac{1-x}{1+

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

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(C) Given equation is $	an ^{-1}\left(\frac{1-x}{1+x}ight)-\frac{1}{2} 	an ^{-1} x=0$. Rearranging the terms,we get $	an ^{-1}\left(\frac{1-x}{1+x}ight)=\frac{1}{2} 	an ^{-1} x$. Using the property $	an ^{-1} a - 	an ^{-1} b = 	an ^{-1}\left(\frac{a-b}{1+ab}ight)$,we can write $	an ^{-1}(1) - 	an ^{-1}(x) = \frac{1}{2} 	an ^{-1} x$. This simplifies to $\frac{\pi}{4} = \frac{1}{2} 	an ^{-1} x + 	an ^{-1} x$. $\frac{\pi}{4} = \frac{3}{2} 	an ^{-1} x$. $	an ^{-1} x = \frac{\pi}{4} 	imes \frac{2}{3} = \frac{\pi}{6}$. Therefore,$x = 	an\left(\frac{\pi}{6}ight) = \frac{1}{\sqrt{3}}$.

If $\tan ^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2} \tan ^{-1} x=0$,for $x>0$,then $x=$

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