中4maths help me!!!急!!

1+2cosθ+cos²θ-sin²θ
=2cosθ+2cos²θ

2010-06-09 23:38:35 補充：

圖片參考：http://h.imagehost.org/0787/ScreenHunter_02_Jun_09_23_37.gif


http://h.imagehost.org/0787/ScreenHunter_02_Jun_09_23_37.gif

L.H.S. = [(1+cosθ) - sinθ] / [(1 + cosθ) + sinθ]
= [(1+cosθ) - sinθ]^2 / {[(1 + cosθ) + sinθ][(1+cosθ) - sinθ]}
= [(1 + cosθ)^2 - 2(1 + cosθ)sinθ + (sinθ)^2] / [(1 + cosθ)^2 - (sinθ)^2]
= [1 + 2cosθ + (cosθ)^2 - 2sinθ - 2sinθcosθ + (sinθ)^2] / [1 + 2cosθ + (cosθ)^2 - (sinθ)^2]
= [2 + 2cosθ - 2sinθ - 2sinθcosθ] / [2cosθ + (cosθ)^2]
= [2(1 + cosθ) - 2sinθ(1 + cosθ)] / [2cosθ(1 + cosθ)]
= [2(1 + cosθ)(1 - sinθ)] / [2cosθ(1 + cosθ)]
= (1 - sinθ) / cosθ
= R.H.S.