Question

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

• A. $\tan x=1$
• B. $\tan 2x=1$
• C. $\tan 3x=1$
• D. None of these

Answer 1

Answer: (C)

$\tan 3x=1$

The given equation can be written as

$\tan x+\frac{\tan x+\tan \left(\frac{\pi }{3}\right)}{1-\tan x\tan \left(\frac{\pi }{3}\right)}+\frac{\tan x+\tan \left(\frac{2\pi }{3}\right)}{1-\tan x\tan \left(\frac{2\pi }{3}\right)}=3$

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

$\Rightarrow \tan x+\frac{(\tan x+\sqrt{3})(1+\sqrt{3}\tan x)+(1-\sqrt{3}\tan x)(\tan x-\sqrt{3})}{1-3{\tan }^{2}x}=3$

$\Rightarrow \tan x+\frac{8\tan x}{1-3{\tan }^{2}x}=3$

$\Rightarrow \frac{\tan x(1-3{\tan }^{2}x)+8\tan x}{1-3{\tan }^{2}x}=3$

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

$\Rightarrow \tan 3x=1$