已知：tan a,tan b是方程X^2+4X+3=0的两个根,求3cos^2(a+b)+sin(a+b)*cos(a+

。。。由韦达定理tan a+tan b=-4tan a*tan b=3tan(a+b)=(tan a+tan b)/(1-tan a*tan b)=2cos^2(a+b)=1/[1+tan^2(a+b)]=1/5sin^2(a+b)=4/5sin(a+b)cos(a+b)=2/5原式=3*1/5+2/5=1