Question

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

Answer 1

$$\begin{gathered} \frac{\pi }{3}=60° \\ \mathrm{LHS}=\tan x+\tan \left(60°+x\right)-\tan \left(60°-x\right) \\ =\tan x+\left(\frac{\tan 60°+\tan x}{1-\tan 60°\tan x}\right)-\left(\frac{\tan 60°-\tan x}{1+\tan 60°\tan x}\right) \\ \left[\tan \left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x\tan y}\mathrm{and}\tan \left(x-y\right)=\frac{\tan x-\tan y}{1+\tan x\tan y}\right] \\ =\tan x+\frac{\sqrt{3}+\tan x}{1-\sqrt{3}\tan x}-\frac{\sqrt{3}-\tan x}{1+\sqrt{3}\tan x} \\ =\tan x+\frac{\sqrt{3}+3\tan x+\tan x+\sqrt{3}{\tan }^{2}x+\sqrt{3}+3\tan x+\tan x-\sqrt{3}{\tan }^{2}x}{\left(1-\sqrt{3}\tan x\right)\left(1+\sqrt{3}\tan x\right)} \\ =\tan x+\frac{8\tan x}{1-3{\tan }^{2}x} \end{gathered}$$

$$\begin{gathered} =\frac{\tan x-3{\tan }^{3}x+8\tan x}{1-3{\tan }^{2}x} \\ =\frac{9\tan x-3{\tan }^{3}x}{1-3{\tan }^{2}x} \\ =3\left(\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}\right)\left(\because \tan 3\mathrm{\theta }=\frac{3\mathrm{tan\theta }-{\tan }^3\mathrm{\theta }}{1-3{\tan }^2\mathrm{\theta }}\right) \\ =3\tan 3x \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$