If $tan^{-1}\frac{1-x}{1+x}=\frac{1}{2}tan^{-1}x$, then the value of x is
Answer & explanation
Correct answer: option 2
We have
$tan^{-1}\frac{1-x}{1+x}=\frac{1}{2}tan^{-1}x$
$⇒ tan^{-1}1-tan^{-1}x =\frac{1}{2}tan^{-1}x $
$⇒ \frac{\pi}{4}=\frac{3}{2}tan^{-1}x ⇒ tan^{-1}x =\frac{\pi}{6}⇒x = \frac{1}{\sqrt{3}}$