(1+sin/1+cos)×(1+sec/1+csc) 求過程?
回答 (2)
2016-09-19 9:47 am
Sol
(1+Sinθ)/(1+Cosθ)*(1+Secθ)/(1+Cscθ)
=[(1+Sinθ)/(1+Cscθ)]/[(1+Secθ)/(1+Cosθ)]
=｛[Sinθ(1+Sinθ)]/[Sinθ(1+Cscθ)]｝/｛[Cosθ(1+Secθ)/[Cosθ(1+Cosθ)]｝
=｛[Sinθ(1+Sinθ)]/(Sinθ+1)｝/｛[(Cosθ+1)/[Cosθ(1+Cosθ)]｝
=Sinθ/Cosθ
=Tanθ
(1+Sinθ)/(1+Cosθ)*(1+Secθ)/(1+Cscθ)
=[(1+Sinθ)/(1+Cscθ)]/[(1+Secθ)/(1+Cosθ)]
=｛[Sinθ(1+Sinθ)]/[Sinθ(1+Cscθ)]｝/｛[Cosθ(1+Secθ)/[Cosθ(1+Cosθ)]｝
=｛[Sinθ(1+Sinθ)]/(Sinθ+1)｝/｛[(Cosθ+1)/[Cosθ(1+Cosθ)]｝
=Sinθ/Cosθ
=Tanθ
2016-09-19 3:21 pm
(1+Sinθ)/(1+Cosθ)*(1+Secθ)/(1+Cscθ)
=[(1+Sinθ)/(1+Cscθ)]/[(1+Secθ)/(1+Cosθ)]
=｛[Sinθ(1+Sinθ)]/[Sinθ(1+Cscθ)]｝/｛[Cosθ(1+Secθ)/[Cosθ(1+Cosθ)]｝
=｛[Sinθ(1+Sinθ)]/(Sinθ+1)｝/｛[(Cosθ+1)/[Cosθ(1+Cosθ)]｝
=Sinθ/Cosθ
=Tanθ
=[(1+Sinθ)/(1+Cscθ)]/[(1+Secθ)/(1+Cosθ)]
=｛[Sinθ(1+Sinθ)]/[Sinθ(1+Cscθ)]｝/｛[Cosθ(1+Secθ)/[Cosθ(1+Cosθ)]｝
=｛[Sinθ(1+Sinθ)]/(Sinθ+1)｝/｛[(Cosθ+1)/[Cosθ(1+Cosθ)]｝
=Sinθ/Cosθ
=Tanθ
收錄日期: 2021-04-18 15:32:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160919004810AACGstb