Question

Prove that:
${\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)-{\tan }^{-1}\left(\frac{4x}{1-4x^2}\right)={\tan }^{-1}2x;|2x|<\frac{1}{\sqrt{3}}$.

Answer 1

${\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)-{\tan }^{-1}\left(\frac{4x}{1-4x^2}\right)={\tan }^{-1}2x$

$\tan ({\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)-{\tan }^{-1}\left(\frac{4x}{1-4x^2}\right))=\tan \left({\tan }^{-1}2x\right)$

$\frac{\frac{6x-8x^3}{1-12x^2}-\frac{4x}{1-4x^2}}{1+\frac{6x-8x^3}{1-12x^2}\cdot \frac{4x}{1-4x^2}}=2x$

$\frac{6x-8x^3}{1-12x^2}-\frac{4x}{1-4x^2}=2x(1+\frac{6x-8x^3}{1-12x^2}\cdot \frac{4x}{1-4x^2})$

$96x^5-6x^4-8x^3+8x^2=0$

$8x^2(12x^3-12x^2-x+1)=0$

$8x^2(x-1)(12x^2-1)=0$

$x=0,1,\pm \frac{1}{2\sqrt{3}}$

$t={\tan }^{-1}\left(\frac{6-8}{1-12}\right)-{\tan }^{-1}\left(\frac{4}{1-4}\right)$

$\tan t=\frac{\frac{2}{11}+\frac{4}{3}}{1+\left(\frac{2}{11}\right)\left(\frac{-4}{3}\right)}$

$=2$