Question

solve for x tan^-1[(1-x)/(1+x)]​

Answer 1

Step-by-step explanation:

We have,

$\tan^{ - 1} ( \frac{1 - x}{1 + x} ) = \frac{1}{2} \tan^{ - 1} (x)$

$let \: \: x = \tan( \alpha )$

$\implies \: \tan^{ - 1} ( \frac{1 - \tan( \alpha ) }{1 + \tan( \alpha ) } ) = \frac{1}{2} \tan^{ - 1} ( \tan( \alpha ) )$

$\implies \tan^{ - 1} ( \tan( \frac{\pi}{4} - \alpha ) ) = \frac{1}{2} \alpha$

$\implies \frac{\pi}{4} - \alpha = \frac{ \alpha }{2}$

$\implies \frac{\pi}{4} = \frac{3 \alpha }{2}$

$\implies \alpha = \frac{\pi}{6}$

so,

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