2x^2-(根号3+1)x+m=0的两根为sinα,cosα,求sinα/(1-cotα)+cosα/(1-tanα)

2x^2-(根号3+1)x+m=0的两根为sinα,cosα,求sinα/(1-cotα)+cosα/(1-tanα)
2x^2-(根号3+1)x+m=0的两根为sinα,cosα,求sinα/(1-cotα)+cosα/(1-tanα)

2x^2-(根号3+1)x+m=0的两根为sinα,cosα,求sinα/(1-cotα)+cosα/(1-tanα)
sinα+cosα=(根号3+1)/2 sinαcosα=m/2
sinα/(1-cotα)+cosα/(1-tanα)
= sinαsinα/( sinα-cosα)+cosαcosα/(cosα-sinα)
=(sinαsinα-cosαcosα)/( sinα-cosα)
=sinα+cosα
=(根号3+1)/2