Question

The value of $\tan x+\tan \left(\frac{\pi }{3}+x\right)+\tan \left(\frac{2\pi }{3}+x\right)$ is
(a) 3 tan 3x
(b) tan 3x
(c) 3 cot 3x
(d) cot 3x

Answer 1

(a) 3 tan 3x

$$\begin{gathered} \frac{\pi }{3}=60°,\frac{2\pi }{3}=120° \\ \tan x+\tan \left(60°+x\right)+\tan \left(120°+x\right)=\tan x+\frac{\tan 60°+\tan x}{1-\tan 60°\tan x}+\frac{\tan 120°+\tan x}{1-\tan 120°\tan x} \\ =\tan x+\frac{\sqrt{3}+\tan x}{1-\sqrt{3}\tan x}+\frac{\left(-\sqrt{3}+\tan x\right)}{1+\sqrt{3}\tan x} \\ =\frac{\tan x\left(1-3{\tan }^{2}x\right)+\left(\sqrt{3}+\tan x\right)\left(1+\sqrt{3}\tan x\right)+\left(-\sqrt{3}+\tan x\right)\left(1-\sqrt{3}\tan x\right)}{1-3{\tan }^{2}x} \\ =\frac{\tan x-3{\tan }^{3}x+\sqrt{3}+3\tan x+\tan x+\sqrt{3}{\tan }^{2}x+\tan x-\sqrt{3}{\tan }^{2}x-\sqrt{3}+3\tan x}{1-3{\tan }^{2}x} \\ =\frac{9\tan x-3{\tan }^{3}x}{1-3{\tan }^{2}x} \\ =\frac{3\left(3\tan x-{\tan }^{3}x\right)}{1-3{\tan }^{2}x} \\ =3\tan 3x \end{gathered}$$