Prove that $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$
Prove that $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$
It is known that $\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$
$\therefore$ $L.H.S.$ $=\tan 4 x=\tan 2(2 x)$
$=\frac{2 \tan 2 x}{1-\tan ^{2}(2 x)}$
$=\frac{2\left(\frac{2 \tan x}{1-\tan ^{2} x}\right)}{1-\left(\frac{2 \tan x}{1-\tan ^{2} x}\right)^{2}}$
$=\frac{\left(\frac{4 \tan x}{1-\tan ^{2} x}\right)}{\left[1-\frac{4 \tan ^{2} x}{\left(1-\tan ^{2} x\right)^{2}}\right]}$
$=\frac{\left(\frac{4 \tan x}{1-\tan ^{2} x}\right)}{\left[\frac{\left(1-\tan ^{2} x\right)^{2}-4 \tan ^{2} x}{\left(1-\tan ^{2} x\right)^{2}}\right]}$
$=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{\left(1-\tan ^{2} x\right)^{2}-4 \tan ^{2} x}$
$=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1+\tan ^{4} x-2 \tan ^{2} x-4 \tan ^{2} x}$
$=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}= R . H.S.$
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