中四數學sin cos tan我要步驟

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http://i841.photobucket.com/albums/zz333/tammie_0666/20100515-3.jpg?t=1273913390

cos^2θ/1-sin^2θ
1+sinθ-cos^2θ/(1+sinθ)cosθ
=1+sinθ-(1-sin^2θ)/(1+sinθ)cosθ
=1-1+sinθ+sin^2θ/(1+sinθ)cosθ
=sinθ+sin^2θ/(1+sinθ)cosθ
=sinθ(1+sinθ)/(1+sinθ)cosθ
=sinθ/cosθ
=tanθ

這是中二的數學嗎？
我現在中二已經教了

2010-05-19 16:38:47 補充：
我也覺得這是中二的數，因為我現在中二已經懂了。

cos^2θ=1 - sin^2θ
1+sinθ - cos^2θ / (1+sinθ)cosθ
=1+sinθ - (1 - sin^2θ)/ (1+sinθ)cosθ
=1-1+sinθ+sin^2θ/(1+sinθ)cosθ
=sinθ+sin^2θ/(1+sinθ)cosθ
=sinθ(1+sinθ)/(1+sinθ)cosθ [[[[抽sinθ出黎]]]
=sinθ/cosθ[[[1+sinθ約左佢]]]
=tanθ[[[sinθ/cosθ=tanθ]]]

2010-05-17 10:15:56 補充：
咁無咁亂
cos^2θ=1 - sin^2θ
1+sinθ - cos^2θ / (1+sinθ)cosθ
=1+sinθ - (1 - sin^2θ)/ (1+sinθ)cosθ
=1-1+sinθ+sin^2θ/(1+sinθ)cosθ
=sinθ+sin^2θ/(1+sinθ)cosθ
=sinθ(1+sinθ)/(1+sinθ)cosθ......................... 抽sinθ出黎
=sinθ/cosθ.....................1+sinθ約左佢
=tanθ............................sinθ/cosθ=tanθ

我想問這不是中二的數嗎??=.=

(1+sinθ-cos^2θ) / [(1+sinθ)cosθ]
= [(1+sinθ)-(1-sinθ)(1+sinθ)]/[(1+sinθ)cosθ]
= [(1+sinθ)(1-1+sinθ)]/[(1+sinθ)cosθ)]
= sinθ/cosθ
= tanθ