Question

Prove the following:
$tan4x=\frac{4tanx(1-ta{n}^{2}x)}{1-6ta{n}^{2}x+ta{n}^{4}x}$

Answer 1

$\tan 4x=\frac{2\tan 2x}{1-{\tan }^{2}2x}$

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

$=\frac{4tanx(1-ta{n}^{2}x)}{{(1-ta{n}^{2}x)}^2-4ta{n}^{2}x}=\frac{4tanx(1-ta{n}^{2}x)}{1-6ta{n}^{2}x+ta{n}^{4}x}$

Hence proved