化简：(cos^2)x/[ctg(x/2)-tg(x/2)]过程详细一点，要快一点谢谢，拜托了！
热心网友
原式=cos^2x/(sinx/1-cosx)-(1-cosx/sinx) =cos^2x/[sin^2x-(1-cosx)^2]/sinx(1-cosx) =cos^2xsinx(1-cosx)/sin^2x-(1-cosx)^2 =cos^2xsinx(1-cosx)/sin^2x-1+2cosx-cos^2x =cos^2xsinx(1-cosx)/2cosx-2cos^2x =cos^2xsinx(1-cosx)/2cosx(1-cosx) =cosxsinx/2=(1/2sin2x)/2=1/4sin2x
热心网友
因为tan(x/2)=[sin(x/2)]/[cos(x/2)]=[2sin(x/2)cos(x/2)]/[2cos(x/2)cos(x/2)]=sinx/(1+cosx),tan(x/2)=[sin(x/2)]/[cos(x/2)]=[2sin(x/2)sin(x/2)]/[2sin(x/2)cos(x/2)]=(1-cosx)/sinx,tanx=1/cotx,所以，(cos^2)x/[ctg(x/2)-tg(x/2)]=(cos^2)x/[(1+cosx)/sinx-(1-cosx)/sinx]=(sinx)(cos^2)x/(2cosx)=sinxcosx/2=(sin2x)/2.