証明恆等式
(a)tanθ．[(1-sinθ)/(1+cosθ)]=cotθ．[(1-cosθ)/(1+sinθ)]
(b)(1-tanθ)²+(1-cotθ)²-(secθ-cosecθ)²=0
a-maths!!!!!
2008-01-29 4:30 am
回答 (2)
2008-01-30 8:19 pm
✔ 最佳答案
(a) 考慮:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jan08/Crazytrigo3.jpg
(b)
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jan08/Crazytrigo4.jpg
參考: My Maths knowledge
2008-01-29 4:46 am
(a) L.H.S/[(1-cosθ)/(1+sinθ)]=tanθ．[(1-sinθ)(1+sinθ)/(1+cosθ)(1-cosθ)]
=tanθ．(1-sin²θ)/(1-cos²θ)=tanθ．cos²θ/sin²θ=cotθ=R.H.S./[(1-cosθ)/(1+sinθ)]
(b)L.H.S=(cosθ-sinθ)²/cos²θ+(cosθ-sinθ)²/sin²θ-(secθ-cosecθ)²
=(cosθ-sinθ)²/cos²θ+(cosθ-sinθ)²/sin²θ-(cosθ-sinθ)²/sin²θcos²θ
=(cosθ-sinθ)²(1/cos²θ+1/sin²θ-1/sin²θcos²θ)
=(cosθ-sinθ)²(cos²θ+sin²θ-1)/sin²θcos²θ
=0
=R.H.S
=tanθ．(1-sin²θ)/(1-cos²θ)=tanθ．cos²θ/sin²θ=cotθ=R.H.S./[(1-cosθ)/(1+sinθ)]
(b)L.H.S=(cosθ-sinθ)²/cos²θ+(cosθ-sinθ)²/sin²θ-(secθ-cosecθ)²
=(cosθ-sinθ)²/cos²θ+(cosθ-sinθ)²/sin²θ-(cosθ-sinθ)²/sin²θcos²θ
=(cosθ-sinθ)²(1/cos²θ+1/sin²θ-1/sin²θcos²θ)
=(cosθ-sinθ)²(cos²θ+sin²θ-1)/sin²θcos²θ
=0
=R.H.S
參考: me
收錄日期: 2021-04-23 20:32:39
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