Question

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

Answer 1

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

$L.H.S=\tan 4x=\tan 2\left(2x\right)=\frac{2\tan 2x}{1-{\tan }^{2}2x}(\because \tan 2A=\frac{2\tan A}{1-{\tan }^{2}A})$

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

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

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

$\Rightarrow L.H.S=R.H.S.$

Hence proved.