Question

tan $x$ + tan$(x+\frac{\pi }{3})+\tan (x+\frac{2\pi }{3})=3⟹\tan 3x=......$

• A. 3
• B. 2
• C. 1
• D. none of these

Answer 1

The correct option is C 1

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

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

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

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

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

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

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

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

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

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

$\Rightarrow$ $\tan 3x=1$

Comparing we get, missing space$=1$