化简:1/(1+sin²α)+1/(1+cos²α)+1/(1+sec²α)+1/(csc
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/16 12:54:18
化简:1/(1+sin²α)+1/(1+cos²α)+1/(1+sec²α)+1/(csc²α)
如题,我实在化不出来
如题,我实在化不出来
![化简:1/(1+sin²α)+1/(1+cos²α)+1/(1+sec²α)+1/(csc](/uploads/image/z/8475972-60-2.jpg?t=%E5%8C%96%E7%AE%80%EF%BC%9A1%2F%281%2Bsin%26%23178%3B%CE%B1%29%2B1%2F%281%2Bcos%26%23178%3B%CE%B1%29%2B1%2F%281%2Bsec%26%23178%3B%CE%B1%29%2B1%2F%28csc)
原式=
1/(1+sin²a)+1/(1+cos²a)+1/[1+(1/cos²a)]+1/[1+(1/sin²a)]
=[1/(1+sin²a)+1/(1+1/sin²a)]+[1/(1+cos²a)+1/(1+1/cos²a)]
=[1/(1+sin²a)+sin²a/(1+sin²a)]+[1/(1+cos²a)+cos²a/(1+cos²a)]
=1+1
=2
1/(1+sin²a)+1/(1+cos²a)+1/[1+(1/cos²a)]+1/[1+(1/sin²a)]
=[1/(1+sin²a)+1/(1+1/sin²a)]+[1/(1+cos²a)+1/(1+1/cos²a)]
=[1/(1+sin²a)+sin²a/(1+sin²a)]+[1/(1+cos²a)+cos²a/(1+cos²a)]
=1+1
=2
化简:1/(1+sin²α)+1/(1+cos²α)+1/(1+sec²α)+1/(csc
(SECα + 1)( SECα - 1)(CSCα + 1)(1 - SINα)
证明,[1+sinα)/(1+cosα)]*[(1+secα)/(1+cscα)]=tanα
sin^2 a/sec^2-1 +cos^2 a/csc^2-1+cos^acsc^a
化简(tanα+tanα*sinα)/(tanα+sinα)*(1+secα)/(1+cscα)
sin cos tan cot sec csc 它们的1度等于?
化简sinθ-cosθ除以(1-tanθ)可得 A -sinθ B -cosθ C -secθ D cscθ
tanα+secα-1/tanα-secα+1=1+sinα/cosα
1/(1+sin^2a)+1/(1+cos^2a)+1/(1+sec^2a)+1/(1+csc^2)
化简(1+cotα-cscα)(1+tanα+secα)
化简:secα√(1+tan^2α)+tanα√(csc^2-1)
化简[(secα-cosα)*(cscα-sinα)]/2sinα*cosα