2x²-（√3+1)x+m=0的两根为 sin θ,cos θ由韦达定理得：sinθ+cosθ=(√3+1)/2sinθcosθ=m/2sinθ/1-cotθ+cosθ/1-tanθ=sin²θ/(sinθ-cosθ)+cos²θ/(cosθ-sinθ)=(sin²θ-cos²θ)/(...sinθ/1-cotθ+cosθ/1-tanθ =sin²θ/(sinθ-cosθ)+cos²θ/(cosθ-sinθ)这个怎么变的？cotθ=cosθ/sinθ；tanθ=sinθ/cosθsinθ/(1-cotθ)+cosθ/(1-tanθ)=sinθ/(1-cosθ/sinθ)+cosθ/(1-sinθ/cosθ)=sinθ/[(sinθ-cosθ)/sinθ]+cosθ/[(cosθ-sinθ)/cosθ]= sin²θ/(sinθ-cosθ)+cos²θ/(cosθ-sinθ)