已知sinαcosα=1/8,且α是第三象限角,求(1-cos^2 α)/(sinα-cosα)-(sinα cosα)/(tan^2 α-1)的值網友廻答：
- 解 化簡(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
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