If $$\tan x +\tan(x+\frac{\pi}{3}) + \tan(x+\frac{2\pi}{3})=3$$ then which of the following is correct?

Substituting $$x+\frac{\pi}{3} = a$$.

$$tan(a-\frac{\pi}{3}) +tan(a)+tan(a+\frac{\pi}{3})=3$$

We know that $$tan(a-b)=\dfrac{tan(a)-tab(b)}{1+tan(a)*tan(b)}$$

Therefore, $$tan(a-\frac{\pi}{3})$$=$$\dfrac{tan(a)-tab(\frac{\pi}{3})}{1+tan(a)*tan(\frac{\pi}{3})}$$

$$tan(a-\frac{\pi}{3})$$=$$\dfrac{tan(a)-\sqrt{3}}{1+\sqrt{3}tan(a)}$$

Similarly, $$tan(a+\frac{\pi}{3})$$=$$\dfrac{tan(a)+\sqrt{3}}{1-\sqrt{3}tan(a)}$$

Hence, $$tan(a-\frac{\pi}{3}) +tan(a)+tan(a+\frac{\pi}{3})$$ = $$\dfrac{tan(a)-\sqrt{3}}{1+\sqrt{3}tan(a)}$$ + tan(a) + $$\dfrac{tan(a)+\sqrt{3}}{1-\sqrt{3}tan(a)}$$

$$\Rightarrow$$ $$\dfrac{8tan(a)}{1-3tan^2(a)}+tan(a)$$

$$\Rightarrow$$ $$\dfrac{9tan(a)-3tan^3(a)}{1-3tan^2(a)}$$

$$\Rightarrow$$ $$3tan(3a)$$

It is given that, $$3tan(3a) = 3$$

Substituting $$a = x+\frac{\pi}{3}$$

$$tan[3(x+\frac{\pi}{3})] = 1$$

$$tan[\pi + 3x] = 1$$ i.e.  $$tan(3x) = 1$$. Hence, option C is the correct answer.