Prove that: \( \tan x \tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right)=\tan3x \)
Solution:
\[
\tan x \tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right)
\]
Using,
\[
\tan\left(\frac{\pi}{3}-x\right)=\frac{\sqrt3-\tan x}{1+\sqrt3\tan x}
\]
and
\[
\tan\left(\frac{\pi}{3}+x\right)=\frac{\sqrt3+\tan x}{1-\sqrt3\tan x}
\]
\[
=\tan x\cdot
\frac{\sqrt3-\tan x}{1+\sqrt3\tan x}
\cdot
\frac{\sqrt3+\tan x}{1-\sqrt3\tan x}
\]
\[
=\tan x\cdot
\frac{3-\tan^2x}{1-3\tan^2x}
\]
\[
=\frac{3\tan x-\tan^3x}{1-3\tan^2x}
\]
Using identity,
\[
\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}
\]
\[
=\tan3x
\]
\[
\boxed{\tan x \tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right)=\tan3x}
\]