令θ＝7π,则有tan3θ+tan4θ＝0,
即tanθ+3tan2θ-3tanθtan²2θ-tan³4θ＝0,
令tanθ＝x,则x^6-21x^4+35x^2-7＝0.
∵tan(kπ/7) (k＝1,2,3)满足上式,
∴tan²(π/7)、tan²(2π/7)、tan²(3π/7)是
y³-21y²+35y-7＝0的根.
依韦达定理知
∴tan²(π/7)tan²(2π/7)tan²(3π/7)＝7
→tan(π/7)tan(2π/7)tan(3π/7)＝√7.
从而,tan(π/7)+tan(2π/7)+tan(4π/7)
＝tan(π/7)+tan(2π/7)-tan(3π/7)
＝tan(3π/7)[1-tan(π/7)tan(2π/7)]-tan(3π/7)
＝-tan(π/7)tan(2π/7)tan(3π/7)
＝-√7.