MHB Simplifying trig expression

[hatelove]

I have to simplify (or get it in terms of tan I guess?) \cot (\frac{2\pi }{3} - x)

I'm not sure how to get the reference angle and subtract the angle 'x' from it to get an expressional value...how would I do this?

 

[soroban]

Hello, daigo!

\text{Simplify: }\:\cot(\tfrac{2\pi}{3} - x)

\text{Identity: }\:\tan(A - B) \;=\;\frac{\tan A - \tan B}{1 + \tan A\tan B}

\cot(\tfrac{2\pi}{3} - x) \;=\;\frac{1}{\tan(\frac{2\pi}{3} - x)} \;=\;\frac{1 + \tan\frac{2\pi}{3}\tan x}{\tan\frac{2\pi}{3} - \tan x}

. . . . . . . . .=\; \frac{1 + (\text{-}\sqrt{3})\tan x}{(\text{-}\sqrt{3}) - \tan x} \;=\; \frac{1 - \sqrt{3}\tan x}{\text{-}\sqrt{3} - \tan x}

. . . . . . . . .=\; \frac{\sqrt{3}\tan x - 1}{\tan x + \sqrt{3}}