Question

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

Answer 1

Given: $\tan x+\tan (x+\frac{\pi }{3})+\tan (x+\frac{2\pi }{3})=3,$

To prove: $\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}=1.$

Now,
$\tan x+\tan (x+\frac{\pi }{3})+\tan (x+\frac{2\pi }{3})=3$

$\Rightarrow \tan x+\frac{\tan x+\tan \frac{\pi }{3}}{1-\tan \frac{\pi }{3}\tan x}+\frac{\tan x+\tan \frac{2\pi }{3}}{1-\tan \frac{2\pi }{3}\tan x}$ $=3$

$\Rightarrow \tan x+\frac{\tan x+\sqrt{3}}{1-\sqrt{3}\tan x}+\frac{\tan x+\tan \frac{2\pi }{3}}{1-\tan \frac{2\pi }{3}\tan x}$ $=3$

$\Rightarrow \tan x+\frac{9\tan x}{1-3\tan x^{2}x}=3$

$\Rightarrow \frac{9\tan x-3\tan x^{2}x}{1-3\tan x^{2}x}=3$

$\therefore \frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}=1.$

Hence proved.