求证:tanα*(1-sinα)/(1+cosα)=cotα*(1-cosα)/(1+sinα)答案上提示说,左边·(tanα·cotα)=左边,但是我还是证不出来~求完整过程.
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![求证:tanα*(1-sinα)/(1+cosα)=cotα*(1-cosα)/(1+sinα)答案上提示说,左边·(tanα·cotα)=左边,但是我还是证不出来~求完整过程.](/uploads/image/z/6866561-65-1.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%9Atan%CE%B1%2A%281-sin%CE%B1%29%2F%281%2Bcos%CE%B1%29%3Dcot%CE%B1%2A%281-cos%CE%B1%29%2F%281%2Bsin%CE%B1%29%E7%AD%94%E6%A1%88%E4%B8%8A%E6%8F%90%E7%A4%BA%E8%AF%B4%2C%E5%B7%A6%E8%BE%B9%C2%B7%EF%BC%88tan%CE%B1%C2%B7cot%CE%B1%EF%BC%89%3D%E5%B7%A6%E8%BE%B9%2C%E4%BD%86%E6%98%AF%E6%88%91%E8%BF%98%E6%98%AF%E8%AF%81%E4%B8%8D%E5%87%BA%E6%9D%A5%7E%E6%B1%82%E5%AE%8C%E6%95%B4%E8%BF%87%E7%A8%8B.)
求证:tanα*(1-sinα)/(1+cosα)=cotα*(1-cosα)/(1+sinα)答案上提示说,左边·(tanα·cotα)=左边,但是我还是证不出来~求完整过程.
求证:tanα*(1-sinα)/(1+cosα)=cotα*(1-cosα)/(1+sinα)
答案上提示说,左边·(tanα·cotα)=左边,但是我还是证不出来~求完整过程.
求证:tanα*(1-sinα)/(1+cosα)=cotα*(1-cosα)/(1+sinα)答案上提示说,左边·(tanα·cotα)=左边,但是我还是证不出来~求完整过程.
tanα*(1-sinα)/(1+cosα)
=[sinα/cosα]*{[(1-sinα)(1-cosα)(1+sinα)]/[(1+cosα)(1+sinα)(1-cosα)]}
=[sinα/cosα]*{[(1-sin²α)(1-cosα)]/[(1-cos²α)(1+sinα)]}
=[sinα/cosα]*{[cos²α(1-cosα)]/[sin²α(1+sinα)]}
=[cosα/sinα]*{(1-cosα)/(1+sinα)}
=cotα*(1-cosα)/(1+sinα)
tanα*(1-sinα)/(1+cosα)=cotα*(1-cosα)/(1+sinα)
<=> tanα*(1-sinα)(1+sinα)=cotα*(1-cosα)(1+cosα)
<=>tanα*(1-sin²α)=cotα*(1-cos²α)
<=>tanα*cos²α=cotα*sin²α
因为tanα=sinα/...
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tanα*(1-sinα)/(1+cosα)=cotα*(1-cosα)/(1+sinα)
<=> tanα*(1-sinα)(1+sinα)=cotα*(1-cosα)(1+cosα)
<=>tanα*(1-sin²α)=cotα*(1-cos²α)
<=>tanα*cos²α=cotα*sin²α
因为tanα=sinα/cosα, cotα=cosα/sinα,
所以,tanα*cos²α=sinαcosα ,cotα*sin²α=sinαcosα
所以tanα*cos²α=cotα*sin²α等式成立
因此tanα*(1-sinα)/(1+cosα)=cotα*(1-cosα)/(1+sinα)等式成立
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