sinα+cosα=tanα
cosα+(cosα)^2/sinα=1
cosα+(1-(sinα)^2)/sinα=1
cosα+1/sinα-sinα-1=0
α∈(0,π/2)時
cosα、1/sinα、-sinα都是減函數(shù)
所以 f(α)=sinα+cosα-tanα是減函數(shù)
f(0)=1>0
f(π/4)=√2-1>0
f(π/3)=√3/2+1/2-√3<0
所以 α∈(π/4,π/3)