§ 23: 23.35

23.35 Доведіть, що коли α, β, γ – кути трикутника, то

tgα2 tgβ2+tgβ2 tgγ2 + tgγ2 tgα2 = 1.

γ = 180° – (α + β), де α, β і γ – кути трикутника, тоді tgγ2 = tg180°(α+ β)2 = tg(90° – α+ β2) = ctgα+ β2 = ctg(α2 + β2) = ctgα2 ctgβ2 -1ctgα2+ ctgβ2 = 1tgα2 • 1tgβ2 -1 1tgα2+1tgβ2 = 1- tgα2 tgβ2tgα2+ tgβ2.

Тоді: tgγ2tgβ2 + tgβ2tgγ2 + tgγ2tgα2 = tgα2tgβ2 + tgγ2(tgβ2 + tgα2) = tgα2tgβ2 + 1- tgα2 tgβ2tgα2+ tgβ2(tgβ2 + tgα2) = tgα2tgβ2 + 1 – tgα2tgβ2 = 1;

що й треба було довести.