Use the identity \(\tan^{-1}\frac{1-x}{1+x} = \tan^{-1}1 - \tan^{-1}x = \frac{\pi}{4} - \tan^{-1}x\) (for \(x > -1\)).

\(\frac{\pi}{4} - \tan^{-1}x = \frac{1}{2}\tan^{-1}x \Rightarrow \frac{3}{2}\tan^{-1}x = \frac{\pi}{4} \Rightarrow \tan^{-1}x = \frac{\pi}{6}\)

\(x = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}\)

Answer: (B) \(\frac{1}{\sqrt{3}}\)