Please answer the following question
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A.Maths Trigonometry
2008-06-08 9:45 pm
回答 (2)
2008-06-08 10:13 pm
✔ 最佳答案
LHS=(cos²x-sinxcosx+tanx)/(cos²x+sinxcosx-tanx)=(cos³x-sinxcos²x+sinx)/(cos³x+sinxcos²x-sinx)
=[cos³x-sinx(1-sin²x)+sinx]/[cos³x+sinx(1-sin²x)-sinx]
=(cos³x+sin³x)/(cos³x-sin³x)
=(1+tan³x)/(1-tan³x) (Divide cos³x on both sides)
=RHS
2008-06-08 10:19 pm
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
=[(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: 2021-04-23 20:33:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080608000051KK01090