设sinθ、cosθ是方程2x^2-(根号3+1)x+m=θ的两根

问题描述:

设sinθ、cosθ是方程2x^2-(根号3+1)x+m=θ的两根
求sinθ/(1-1/tanθ)+cosθ/(1-tanθ)的值 m=根号3/2

sinθ、cosθ是方程2x^2-(根号3+1)x+m=θ的两根;因此
sinθ+cosθ=(√3+1)/2
化简sinθ/(1-1/tanθ)+cosθ/(1-tanθ)得到sinθ/(1-1/tanθ)+cosθ/(1-tanθ)=sinθ+cosθ
=(√3+1)/2我要化简的过程sinθ/(1-1/tanθ)+cosθ/(1-tanθ)=sinθ/(1-cosθ/sinθ)+cosθ/(1-sinθ/cosθ)=sin^2 θ/(sinθ-cosθ)+cos^2 θ/(cosθ-sinθ)=sin^2 θ/(sinθ-cosθ)-cos^2 θ/(sinθ-cosθ)=(sin^2 θ-cos^2 θ)/(sinθ-cosθ)=(sinθ+cosθ)(sinθ-cosθ)/(sinθ-cosθ)=sinθ+cosθ