Question

已知（2sin^2α+2sinαcosα)/(1+tanα)=k(0

Answer 1

(2sin^2 α+2sinα*cosα)/(1+tanα)=k (2sin^2 α+2sinα*cosα)/(1+tanα)=2sinacosa(sina+cosa)/(sina+cosa)=2sinacosa=k (sinα-cosα)^2=1-2sinacosa=1-k sinα-cosα=±√（1-k）