Question

If tan x + $\tan \left(x+\frac{\mathrm{\pi }}{3}\right)+\tan \left(x+\frac{2\mathrm{\pi }}{3}\right)=3$, then prove that $\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}=1$.

Answer 1

$$\begin{gathered} \text{Given:} \\ \tan x+\tan \left(x+\frac{\mathrm{\pi }}{3}\right)+\tan \left(\mathrm{x}+\frac{2\mathrm{\pi }}{3}\right)=3 \\ \Rightarrow \tan x+\frac{\tan x+\tan {\displaystyle \frac{\mathrm{\pi }}{3}}}{1-\tan x\tan {\displaystyle \frac{\mathrm{\pi }}{3}}}+\frac{\tan x+\tan {\displaystyle \frac{2\mathrm{\pi }}{3}}}{1-\tan x\tan {\displaystyle \frac{2\mathrm{\pi }}{3}}}=3 \\ \Rightarrow \tan x+\frac{\tan x+\sqrt{3}}{1-\sqrt{3}\tan x}+\frac{\tan x-\sqrt{3}}{1+\sqrt{3}\tan x}=3\left[\tan 120°=-\sqrt{3}\right] \\ \Rightarrow \frac{\tan x(1-3{\tan }^{2}x)+\tan x+\sqrt{3}+\sqrt{3}{\tan }^{2}x+3\tan x+\tan x-\sqrt{3}-\sqrt{3}{\tan }^{2}x+3\tan x}{1-3{\tan }^{2}x}=3 \\ \Rightarrow \frac{9\tan x-3{\tan }^{3}x}{1-3{\tan }^{2}x}=3 \\ \Rightarrow \frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}=1 \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$