Prove the identity
[(cosθ)^2-sinθcosθ+tanθ)]/[(cosθ)^2+sinθcosθ-tanθ]
=[1+(tanθ)^3]/[1-(tanθ)^3]
Prove identity
2008-02-09 12:54 am
回答 (1)
2008-02-09 1:06 am
✔ 最佳答案
LHS=[(cosθ)^2-sinθcosθ+tanθ)]/[(cosθ)^2+sinθcosθ-tanθ]
=[(cosθ)^3-sinθcos^2θ+sinθ)]/[(cosθ)^3+sinθcos^2θ-sinθ]
=[(cosθ)^3-sinθ(cos^2θ-1)]/[(cosθ)^3+sinθ(cos^2θ-1)]
=[(cosθ)^3+sin^3θ]/[(cosθ)^3-sin^3θ]
=[1+(tanθ)^3]/[1-(tanθ)^3]
=RHS
收錄日期: 2021-04-25 16:54:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080208000051KK01513