proof left as an exercise

[Albert Chan]

Another proof, letting t = tan(10°)

tan(30°) = 1/√3 = (3t-t³) / (1-3t²)                  // triple-angle formula
⇒ t³ = √3*t² + 3t - 1/√3
⇒ t⁴ = 6t² + 8/√3*t - 1

2*cos(30°) / (1 + 4*cos(20°)) ?=? tan(20°)

√3 / (1 + 4*(1-t²)/(1+t²)) ?=? 2t / (1-t²)        // tangent half-angle formula
√3*(1+t²) / (5-3t²) ?=? 2t / (1-t²)                  // cross multiply
√3*(1-t⁴) ?=? 10t - 6t³

√3*(1-t⁴) + 6t³ - 10t
= √3 - √3*(6t² + 8/√3*t - 1) + 6*(√3*t² + 3t - 1/√3) - 10t
= 0      QED