Question

$\tan x\tan \left(x+\frac{\pi }{3}\right)+\tan x\tan \left(\frac{\pi }{3}-x\right)+\tan \left(x+\frac{\pi }{3}\right)\tan \left(x-\frac{\pi }{3}\right)=-3$

Answer 1

$$\begin{gathered} \frac{\pi }{3}=60° \\ \mathrm{LHS}=\tan x\tan \left(x+60°\right)+\tan x\tan \left(x-60°\right)+\tan \left(x+60°\right)\tan \left(x-60°\right) \\ =\tan x\left(\frac{\tan x+\sqrt{3}}{1-\sqrt{3}\tan x}\right)+\tan x\left(\frac{\tan x-\sqrt{3}}{1+\sqrt{3}\tan x}\right)+\left(\frac{\tan x+\sqrt{3}}{1-\sqrt{3}\tan x}\right)\left(\frac{\tan x-\sqrt{3}}{1+\sqrt{3}\tan x}\right) \\ \left[\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}\mathrm{and}\tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}\right] \\ =\frac{\left(1+\sqrt{3}\tan x\right)\tan x\left(\tan x+\sqrt{3}\right)+\left(1-\sqrt{3}\tan x\right)\tan x\left(\tan x-\sqrt{3}\right)+{\tan }^{2}x-3}{\left(1+\sqrt{3}\tan x\right)\left(1-\sqrt{3}\tan x\right)} \\ =\frac{\left(\tan x+\sqrt{3}{\tan }^{2}x\right)\left(\tan x+\sqrt{3}\right)+\left(\tan x-\sqrt{3}\tan x\right)\left(\tan x-\sqrt{3}\right)+{\tan }^{2}x-3}{1-3{\tan }^{2}x} \end{gathered}$$