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

Answer 1

(c) tan 3x

$$\begin{gathered} \frac{\pi }{3}=60° \\ \tan x\tan \left(60°-x\right)\tan \left(60°+x\right)=\tan x\times \frac{\tan 60°-\tan x}{1+\tan 60°\tan x}\times \frac{\tan 60°+\tan x}{1-\tan 60°\tan x} \\ =\tan x\times \frac{\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}\times \frac{\sqrt{3}+\tan x}{1-\sqrt{3}\tan x} \\ =\frac{\tan x\left(3-{\tan }^{2}x\right)}{1-3{\tan }^{2}x} \\ =\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x} \\ =\tan 3x \end{gathered}$$