Question

Prove that : $ta{n}^{-1}\left[\frac{6x-8x^3}{1-12x^2}\right]-ta{n}^{-1}\left[\frac{4x}{1-4x^2}\right]=ta{n}^{-1}2x,\left|2x\right|<\frac{1}{\sqrt{3}}$

Answer 1

${\tan }^{-1}\left[\frac{6x-8x^3}{1-12x^2}\right]-{\tan }^{-1}\left[\frac{4x}{1-4x^2}\right]={\tan }^{-1}2x,|2x|<\frac{1}{\sqrt{3}}$

On solving LHS we get ${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$

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

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

${\tan }^{-1}\left[\frac{6x-24x^3-8x^3+32x^5-4x+48x^3}{1-4x^2-12x^2+48x^4+24x^2-32x^4}\right]$

${\tan }^{-1}\left[\frac{32x^5+16x^3+2x}{16x^4+8x^2+1}\right]$

${\tan }^{-1}\left[\frac{2x[16x^4+8x^2+1]}{16x^4+8x^2+1}\right]$

${\tan }^{-1}\left(2x\right)=$RHS

Hence proved.