Question

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

Answer 1

${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$

$\Rightarrow 3{\tan }^{-1}x={\tan }^{-1}x+{\tan }^{-1}x+{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\tan }^{-1}x$

$={\tan }^{-1}\left(\frac{\frac{2x}{1-x^2}+x}{1-\frac{x-2x}{1-x^2}}\right)={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)=$R.H.S.