已知tan(α+β)=-2,tan(α-β)=1/2,求sin2α/sin2β的值

tan(α+β)=sin(α+β)/cos(α+β)=-2 (1)
tan(α-β)=sin(α-β)/cos(α-β)=1/2 (2)
(1)/(2)
sin(α+β)cos(α-β)/cos(α+β)sin(α-β)=-4
sin2α/sin2β=sin[(α+β)+(α-β)]/sin[(α+β)-(α-β)]
=[sin(α+β)cos(α-β)+cos(α+β)sin(α-β)]/[sin(α+β)cos(α-β)-cos(α+β)sin(α-β)](上下同除以cos(α+β)sin(α-β))
=-4+1/(-4-1)
=3/5