方程x^2-3x-3两根tana tanβ 求sin(α+β)^2-cos(α+β)^2

tana+tanb=3
tanatanb=-3
tan(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=16/25
sin^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