求证:(tanα+secα-1)/(tanα-secα+1)=(secα+tanα+1)/(secα-tanα+1)
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求证:(tanα+secα-1)/(tanα-secα+1)=(secα+tanα+1)/(secα-tanα+1)
![求证:(tanα+secα-1)/(tanα-secα+1)=(secα+tanα+1)/(secα-tanα+1)](/uploads/image/z/17959449-57-9.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BC%88tan%CE%B1%2Bsec%CE%B1-1%EF%BC%89%2F%28tan%CE%B1-sec%CE%B1%2B1%29%3D%28sec%CE%B1%2Btan%CE%B1%2B1%EF%BC%89%2F%28sec%CE%B1-tan%CE%B1%2B1%EF%BC%89)
原式:
![](http://img.wesiedu.com/upload/6/dd/6ddf3201ae31a72f069e2538b8c37c3d.jpg)
求证:(1+secα+tanα)/(1+secα+tanα)=(1+sinα)/cosα
tanα+secα-1/tanα-secα+1=1+sinα/cosα
求证:(tanα+secα-1)/(tanα-secα+1)=(1+sinα)/cosα
求证1+sinα/cosα=tanα+secα-1/tanα-secα+1
(tanα+cotα)/ (secα*cscα)=?
化简:secα√(1+tan^2α)+tanα√(csc^2-1)
化简(tanα+tanα*sinα)/(tanα+sinα)*(1+secα)/(1+cscα)
tanα+cotα=secα·cscα
证明,[1+sinα)/(1+cosα)]*[(1+secα)/(1+cscα)]=tanα
求证1+tan^2α=sec^2α,1+cot^2α=csc^2α
三角函数 sec(2a)=sec^a/1-tan^a
求证明三角函数等式啊 1.tanα+cosα/(1+sinα)=secα 2.cotα/(1-tanα)+tanα/(1