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Question
If tan x + \[\tan \left( x + \frac{\pi}{3} \right) + \tan \left( x + \frac{2\pi}{3} \right) = 3\], then prove that \[\frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} = 1\].
Solution
\[\tan x + \tan\left( x + \frac{\pi}{3} \right) + \tan\left( x + \frac{2\pi}{3} \right) = 3\]
\[ \Rightarrow \tan x + \frac{\tan x + \tan\frac{\pi}{3}}{1 - \tan x \tan \frac{\pi}{3}} + \frac{\tan x + \tan\frac{2\pi}{3}}{1 - \tan x \tan\frac{2\pi}{3}} = 3\]
\[ \Rightarrow \tan x + \frac{\tan x + \sqrt{3}}{1 - \sqrt{3}\tan x} + \frac{\tan x - \sqrt{3}}{1 + \sqrt{3}\tan x} = 3 \left[ \tan120^\circ = - \sqrt{3} \right]\]
\[ \Rightarrow \frac{\tan x(1 - 3 \tan^2 x) + \tan x + \sqrt{3} + \sqrt{3} \tan^2 x + 3\tan x + \tan x - \sqrt{3} - \sqrt{3} \tan^2 x + 3\tan x}{1 - 3 \tan^2 x} = 3 \]
\[ \Rightarrow \frac{9\tan x - 3 \tan^3 x}{1 - 3 \tan^2 x} = 3\]
\[ \Rightarrow \frac{3\tan x - \tan^3 x}{1 - 3 \tan^2 x} = 1\]
Hence proved .
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