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

solution:

It is known that $\tan 2 \mathrm{~A}=\frac{2 \tan \mathrm{A}}{1-\tan ^{2} \mathrm{~A}}$.

$\therefore$ L.H.S. $=\tan 4 x=\tan 2(2 x)$

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

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

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

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

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

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

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