Пусть
arctg(x+1)=α; ⇒ tgα=x+1; α∈(-π/2;π/2)
arctg(x-1)=β ⇒ tgβ=x-1; β∈(-π/2;π/2)
Уравнение:
α+β=arctg2
tg(α+β)=tg(arctg2)
tg(α+β)=2
при х=1
arctg2+arctg0=arctg2 - верно
при х=-2
arctg(-1)+arctg(-3)=arctg2 - неверно
О т в е т. 1
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Verified answer
Пусть
arctg(x+1)=α; ⇒ tgα=x+1; α∈(-π/2;π/2)
arctg(x-1)=β ⇒ tgβ=x-1; β∈(-π/2;π/2)
Уравнение:
α+β=arctg2
tg(α+β)=tg(arctg2)
tg(α+β)=2
при х=1
arctg2+arctg0=arctg2 - верно
при х=-2
arctg(-1)+arctg(-3)=arctg2 - неверно
О т в е т. 1