2x^2-(根号3+1)x+m=0的两根为sinα,cosα,求sinα/(1-cotα)+cosα/(1-tanα)
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![2x^2-(根号3+1)x+m=0的两根为sinα,cosα,求sinα/(1-cotα)+cosα/(1-tanα)](/uploads/image/z/805704-24-4.jpg?t=2x%5E2-%28%E6%A0%B9%E5%8F%B73%2B1%29x%2Bm%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E4%B8%BAsin%CE%B1%2Ccos%CE%B1%2C%E6%B1%82sin%CE%B1%2F%281-cot%CE%B1%29%2Bcos%CE%B1%2F%281-tan%CE%B1%29)
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