Question:
The value of $\tan x \tan \left(\frac{\pi}{3}-x\right) \tan \left(\frac{\pi}{3}+x\right)$ is
(a) cot 3x
(b) 2cot 3x
(c) tan 3x
(d) 3 tan 3x
Solution:
(c) tan 3x
$\frac{\pi}{3}=60^{\circ}$
$\tan x \tan \left(60^{\circ}-x\right) \tan \left(60^{\circ}+x\right)=\tan x \times \frac{\tan 60^{\circ}-\tan x}{1+\tan 60^{\circ} \tan x} \times \frac{\tan 60^{\circ}+\tan x}{1-\tan 60^{\circ} \tan x}$
$=\tan x \times \frac{\sqrt{3}-\tan x}{1+\sqrt{3} \tan x} \times \frac{\sqrt{3}+\tan x}{1-\sqrt{3} \tan x}$
$=\frac{\tan x\left(3-\tan ^{2} x\right)}{1-3 \tan ^{2} x}$
$=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}$
$=\tan 3 x$