高中三角函数问题:已知tan(α+β)=2/5,tan(β-π/4)=1/4,则tan(α+π/4)的值为多少?
(α+π/4)=(α+β)-(β-π/4)
tan(α+π/4)
=[tan(α+β)-tan(β-π/4)]/[1+tan(α+β)tan(β-π/4)]
=(2/5-1/4)/[1+(2/5)*(1/4)]
=(3/20)/(11/10)
=3/22
高中三角函数问题:已知tan(α+β)=2/5,tan(β-π/4)=1/4,则tan(α+π/4)的值为多少?
(α+π/4)=(α+β)-(β-π/4)
tan(α+π/4)
=[tan(α+β)-tan(β-π/4)]/[1+tan(α+β)tan(β-π/4)]
=(2/5-1/4)/[1+(2/5)*(1/4)]
=(3/20)/(11/10)
=3/22