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Question: If \(\tan x + \tan\left( \frac{\pi}{3} + x \right) + \tan\left( \frac{2\pi}{3} + x \right) = 3,\) th...

If tan⁡x+tan⁡(π3+x)+tan⁡(2π3+x)=3,\tan x + \tan\left( \frac{\pi}{3} + x \right) + \tan\left( \frac{2\pi}{3} + x \right) = 3, then

A

tan⁡x=1\tan x = 1

B

tan⁡2x=1\tan 2x = 1

C

tan⁡3x=1\tan 3x = 1

D

None of these

Answer

tan⁡3x=1\tan 3x = 1

Explanation

Solution

tan⁡x+tan⁡(π3+x)+tan⁡(2π3+x)\tan x + \tan\left( \frac{\pi}{3} + x \right) + \tan\left( \frac{2\pi}{3} + x \right)

=tan⁡x+tan⁡x+31−3tan⁡x+tan⁡x−31+3tan⁡x= \tan x + \frac{\tan x + \sqrt{3}}{1 - \sqrt{3}\tan x} + \frac{\tan x - \sqrt{3}}{1 + \sqrt{3}\tan x}

=tan⁡x+8tan⁡x1−3tan⁡2x=3(3tan⁡x−tan⁡3x)1−3tan⁡2x=3tan⁡3x= \tan x + \frac{8\tan x}{1 - 3\tan^{2}x} = \frac{3(3\tan x - \tan^{3}x)}{1 - 3\tan^{2}x} = 3\tan 3x

Therefore, the given equation is 3tan⁡3x=33\tan 3x = 3⇒ tan⁡3x=1.\tan 3x = 1.