Question

已知tanα,tanβ是方程x^2+4x-3=0的两个根,求sin^2（α+β）+sin(α+β)cos(α+β)=

Answer 1

tanα+tanβ=-4

tanαtanβ=-3

tan(α+β)=-4/(1+3)=-1

sin^2（α+β）+sin(α+β)cos(α+β)

=[sin^2（α+β）+sin(α+β)cos(α+β)]/1

=[sin^2（α+β）+sin(α+β)cos(α+β)]/[sin^2（α+β）+cos^2(α+β)]

=[tan^2(α+β)+tan(α+β)]/(1+tan^2(α+β))

=(1-1)/(1+1)

=0