tanx=-1/2,sin2x+cos2x/4cos^2x-3sin^2x+1=

(1)、ab=0,2cosx-sinx=0,4cos x=sin x,cos x=1/5,f(x)=(1+sin2x+cos2x)/(1+tanx.)=(2sinxcosx+2cos x)/(1+tanx.) =2cos x(sinx

sin2x+cos2x/4cos^2x-3sin^2x+1=sin2x+cos2x/4cos^2x-3+3cos^2x+1
=sin2x+cos2x/7cos^2x-2=sin2x+cos2x/7cos^2x-7/2+3/2=sin2x+cos2x/7cos2x/2+3/2
根据万能公式
sin2x=2tanx/(1+tanx^2)=-1/(5/4)=-4/5
cos2x=(1-tanx^2)/(1+tanx^2)=(3/4)/(5/4)=3/5
原式=(-4/5+3/5)/(7/2*3/5+3/2)=(-1/5)/(18/5)=-1/18