Question

Prove that ${tan}^{-1}x+{tan}^1\left(\frac{2x}{1-x^2}\right)={tan}^{-1}\left(\frac{3x-x^3}{1-{3x}^2}\right)$ for $\left|x\right|<\frac{1}{3}$.

Answer 1

$LHS=ta{n}^{-1}x+ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)$

we know, $ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+y}{1=xy}\right)$

$ta{n}^{-1}\left[\frac{x+\frac{2x}{1-x2}}{1-\frac{2x2}{1-x2}}\right]]$

$ta{n}^{-1}\left[\frac{x-x^3+2x}{1-x^2-2x^2}\right]$

$ta{n}^{-1}\left[\frac{3x-x^3}{1-3x^2}\right]\left|x\right|<1/3$