Question

If $\tan^{-1}\frac{1-x}{1+x}=\frac{1}{2}\tan^{-1}x$, then the value of $x$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt3}$
• C. $\sqrt3$
• D. $2$

Answer 1

Answer: (B)

$\frac{1}{\sqrt3}$

Answer 2

Put $\frac{1}{2}tan^{-1} x = \theta$ $\Rightarrow x = tan\, 2\,\theta$ $\therefore \frac{1-x}{1+x} = \frac{1-tan\,2\,\theta}{1+tan\,2\,\theta}$ $= tan\left(\frac{\pi}{4}-2\,\theta\right)$ $\therefore tan^{-1} \frac{1-x}{1+x} = \frac{\pi}{4} -2\,\theta$ $\therefore \frac{\pi}{4} -2\,\theta = \theta$ $\Rightarrow 3\,\theta = \frac{\pi}{4}$ $\Rightarrow \theta = \frac{\pi}{12}$ $\Rightarrow 2\,\theta= \frac{\pi}{6}$ $\therefore x= tan\, 2\,\theta$ $= tan \frac{\pi}{6}$ $= \frac{1}{\sqrt{3}}$