Question

Solve:
$y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

Answer 1

Given that,

$y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

Put $x=\tan \theta$ then, $\theta ={\tan }^{-1}x$

$y={\tan }^{-1}\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)$

$y={\tan }^{-1}\tan 3\theta$

$y=3\theta$

$y=3{\tan }^{-1}x,-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$

Hence, this is the answer.