Question

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

Answer 1

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

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

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

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

Hence proved