tanA=2 求(sin^2 A+ sinAcosA)/(tanA+1+sin^2 A) 急 .
问题描述：
tanA=2 求(sin^2 A+ sinAcosA)/(tanA+1+sin^2 A) 急 .
答
∵tanA=2
∴sin2A=2tanA/(1+tan^2 A)=4/5
cos2A=(1-tan^2 A)/(1+tan^2 A)=-3/5
∴原式=[(1-cos2A)/2+sin2A/2]/(tanA+3/2-cos2A/2)
=6/19