Question

Prove that: $\tan 4x=\frac{4\tan x(1-{\tan }^{2}x)}{1-6{\tan }^{2}x+{\tan }^{4}x}$

Answer 1

We know that, $tan2A=\frac{2tanA}{1-{tan}^{2}A}$

$\therefore \tan 4x=\tan 2\left(2x\right)$ $=\frac{2\tan 2x}{1-{\tan }^{2}2x}$

$=\frac{2\times \frac{2\tan x}{1-{\tan }^{2}x}}{1-{\left[\frac{2\tan x}{1-{\tan }^{2}x}\right]}^2}$

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

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

$=\frac{4\tan x\left(1-{\tan }^{2}x\right)}{1+{\tan }^{4}x-2{\tan }^{2}x-4{\tan }^{2}x}$

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