GalaxyXy
幼苗
共回答了25个问题采纳率：88% 举报
左边=(1-cos²α)/(sinα-cosα)-(sinα+cosα)/(tan^2α-1)
=sin²α/(sinα-cosα)-(sinα+cosα)/(sin²a/cos²α-1)
=sin²α/(sinα-cosα)-(sinα+cosα)cos²α/(sin²α-cos²α)
=sin²α/(sinα-cosα)-cos²α/(sinα-cosα)
=(sin²α-cos²α)/(sina-cosα)
=(sinα+cosα)(sinα-cosα)/(sinα-cosα)
=sinα+cosα=右边
∴等式成立
1年前
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