If $\tan x + \tan (x + π / 3) +$ $\tan (x + 2 π / 3) = 3$ then

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$\tan 3 x = 1$

$\tan 2 x = 1$

none of these.

$\tan x = 1$

answer is C.

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Detailed Solution

The given equation can be written as

$\begin{array}{l} \Rightarrow \tan x + \frac{\tan x + \tan (π / 3)}{1 - \tan x \tan (π / 3)} + \frac{\tan x + \tan (2 π / 3)}{1 - \tan x \tan (2 π / 3)} = 3 \\ \Rightarrow \tan x + \frac{(\tan x + \sqrt{3}) (1 + \sqrt{3} \tan x) (1 - \sqrt{3} \tan x) (\tan x - \sqrt{3})}{1 - 3 \tan^{2} x} = 3 \\ \Rightarrow \tan x + \frac{8 \tan x}{1 - 3 \tan^{2} x} = 3 \end{array}$

$\begin{array}{l} \Rightarrow \frac{\tan x (1 - 3 \tan^{2} x) + 8 \tan x}{1 - 3 \tan^{2} x} = 3 \\ \Rightarrow \frac{3 (3 \tan x - \tan^{3} x)}{1 - 3 \tan^{2} x} = 3 \Rightarrow 3 \tan 3 x = 3 \\ \Rightarrow \tan 3 x = 1 . \end{array}$