Question

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

• A. 1/2
• B.
  
• C.
  
• D. 2

Answer 1

The correct option is B

$\because ta{n}^{-1}\frac{1-x}{1+x}=\frac{1}{2}ta{n}^{-1}x$
Let $x=tan\theta$
$\therefore ta{n}^{-1}\left(\frac{1-tan\theta }{1+tan\theta }\right)=\frac{1}{2}ta{n}^{-1}\left(tan\theta \right)$
$\Rightarrow ta{n}^{-1}\left(tan(\frac{\pi }{4}-\theta )\right)=\frac{1}{2}ta{n}^{-1}\left(tan\theta \right)$
$\Rightarrow \frac{\pi }{4}-\theta =\frac{\theta }{2}$
$\Rightarrow \frac{3\pi }{2}=\frac{\pi }{4}\Rightarrow \theta =\frac{\pi }{6}$
$\therefore x=tan\theta =tan\frac{\pi }{6}=\frac{1}{\sqrt{3}}$