Solve tan ^{-1} frac{1-x}{1+x}=frac{1}{2} tan | vedclass

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(B) Given the equation: $	an ^{-1} \frac{1-x}{1+x}=\frac{1}{2} 	an ^{-1} x$Using the identity $	an ^{-1} a - 	an ^{-1} b = 	an ^{-1} \frac{a-b}{1

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

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(B) Given the equation: $	an ^{-1} \frac{1-x}{1+x}=\frac{1}{2} 	an ^{-1} x$Using the identity $	an ^{-1} a - 	an ^{-1} b = 	an ^{-1} \frac{a-b}{1+ab}$,we can write $	an ^{-1} 1 - 	an ^{-1} x = \frac{1}{2} 	an ^{-1} x$.Since $	an ^{-1} 1 = \frac{\pi}{4}$,the equation becomes $\frac{\pi}{4} - 	an ^{-1} x = \frac{1}{2} 	an ^{-1} x$.Rearranging the terms,we get $\frac{\pi}{4} = 	an ^{-1} x + \frac{1}{2} 	an ^{-1} x = \frac{3}{2} 	an ^{-1} x$.Multiplying both sides by $\frac{2}{3}$,we get $	an ^{-1} x = \frac{\pi}{4} 	imes \frac{2}{3} = \frac{\pi}{6}$.Taking the tangent of both sides,$x = 	an \frac{\pi}{6}$.Therefore,$x = \frac{1}{\sqrt{3}}$.

Solve $\tan ^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan ^{-1} x$ for $x > 0$.

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