已知tanα、tanβ是方程x^2-3x-3=0的两根,求sin^2(α+β)-3sin(α+β)cos(α+β)-3c

tana+tanb=3tanatanb=-3tan(a+b)=3/[1-(-3)]=3/4=sin(a+b)/cos(a+b)sin(a+b)=(3/4)*cos(a+b)[sin(a+b)]^2+[cos(a+b)]^2=1[cos(a+b)]^2=16/25sin^2(α+β)-3sin(α+β)cos(α+β)-3cos^2(α+β)=cos^2(a+b)[tan^2(a+b)-3tan(a+b)-3]=(16/25)[9/16-9/4-3]=-3