已知(tanα-1)/tanα=-1,求1/(sin²α+2sinαcosα-cos²α)的值

问题描述:

已知(tanα-1)/tanα=-1,求1/(sin²α+2sinαcosα-cos²α)的值

(tanα-1)/tanα=-1tanα-1=-tanα2tanα=1tanα=1/21/(sin²α+2sinαcosα-cos²α)=(sin²α+cos²α)/(sin²α+2sinαcosα-cos²α)=(tan²α+1)/(tan²α+2tanα-1)=[(1/2)...