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

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

已知(tanα-1)/tanα=-1,求1/(sin²α+2sinαcosα-cos²α)的值
(tanα-1)/tanα=-1
tanα-1=-tanα
2tanα=1
tanα=1/2
1/(sin²α+2sinαcosα-cos²α)
=(sin²α+cos²α)/(sin²α+2sinαcosα-cos²α)
=(tan²α+1)/(tan²α+2tanα-1)
=[(1/2)²+1]/[(1/2)²+2(1/2)-1]
=(5/4)/(1/4)
=5