Solve for x: tan^{-1}left(frac{1-x}{1+x}right | vedclass

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(D) Given the equation: $	an^{-1}\left(\frac{1-x}{1+x}ight) = \frac{1}{2} 	an^{-1} x$.We know that $	an^{-1}\left(\frac{1-x}{1+x}ight) = 	an^{

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

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(D) Given the equation: $	an^{-1}\left(\frac{1-x}{1+x}ight) = \frac{1}{2} 	an^{-1} x$.We know that $	an^{-1}\left(\frac{1-x}{1+x}ight) = 	an^{-1}(1) - 	an^{-1}(x)$.Since $	an^{-1}(1) = \frac{\pi}{4}$,the equation becomes: $\frac{\pi}{4} - 	an^{-1} x = \frac{1}{2} 	an^{-1} x$.Adding $	an^{-1} x$ to both sides: $\frac{\pi}{4} = \frac{1}{2} 	an^{-1} x + 	an^{-1} x$.$\frac{\pi}{4} = \frac{3}{2} 	an^{-1} x$.$	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}}$.

Solve for $x$: $\tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2} \tan^{-1} x$,where $x > 0$.

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