MHB Simplifying trig expression

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To simplify the expression cot(2π/3 - x), the identity tan(A - B) = (tan A - tan B) / (1 + tan A tan B) is applied. The cotangent is rewritten in terms of tangent, leading to cot(2π/3 - x) = 1/tan(2π/3 - x). Substituting tan(2π/3) = -√3 into the equation results in cot(2π/3 - x) being expressed as (√3 tan x - 1) / (tan x + √3). The final simplified form of the expression is achieved as (√3 tan x - 1) / (tan x + √3). This method effectively transforms the original cotangent expression into a more manageable form.
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I have to simplify (or get it in terms of tan I guess?) \cot (\frac{2\pi }{3} - x)

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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?
 
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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}}
 
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