已知0＜α＜π/2,且3sinα=4cosα求(sin^2α+2sinαcosα)/(3cos^2α-1)求cos^2α+sinαcosα

因为：3sina=4cosa,所以：tana=4/3(1)(sin^2α+2sinαcosα)/(3cos^2α-1)=(sin^2α+2sinαcosα)/(2cos^2α-sin^2a)=(tan^2a+2tana)/(2-tan^2)=(16/9+8/3)/(2-16/9)=20(2)cos^2α+sinαcosα =(cos^2α+sinαcosα ...