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已知(tanα-1)/tanα=-1,求1/(sin²α+2sinαcosα-cos²α)的值
人气:338 ℃ 时间:2020-10-02 00:54:55
解答
(tanα-1)/tanα=-1tanα-1=-tanα2tanα=1tanα=1/21/(sin²α+2sinαcosα-cos²α)=(sin²α+cos²α)/(sin²α+2sinαcosα-cos²α)=(tan²α+1)/(tan²α+2tanα-1)=[(1/2)&#...
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