查詢關於三角學的問題

cos^2θ+sin^2θ=1

1. 1/cos^2θ－(1+tan^2θ)
= (1/cos^2θ－(1+tan^2θ)) * (cos^2θ /cos^2θ)
= (1 - (cos^2θ+sin^2θ) ) /cos^2θ
= (1-1)/cos^2θ
= 0

2. (sinθ+cosθ)^2－1
= cos^2θ+sin^2θ+2sinθcosθ -1
= 2sinθcosθ

3. 係咪cosθ/(1 + sinθ) － 1－sinθ/cosθ??

2007-03-25 03:21:03 補充：
....好彩冇計到, 計計下唔對路, 所以至問下....(cosθ)/(1 sinθ)-(1-sinθ)/(cosθ)=(cos^2θ - (1 sinθ)*(1-sinθ) ) / ((1 sinθ)cosθ) ...通母分=(cos^2θ - (1-sin^2θ) ) / ((1 sinθ)cosθ)= (cos^2θ sin^2θ -1) / ((1 sinθ)cosθ)=0

2007-03-25 03:24:29 補充：
(cosθ)/(1 ＋ sinθ)-(1-sinθ)/(cosθ)=[cos^2θ - (1 sinθ)*(1-sinθ) ] / [(1 ＋ sinθ)*cosθ] ...通母分=[cos^2θ - (1-sin^2θ) ] / [(1 ＋ sinθ)*cosθ]= [cos^2θ＋sin^2θ -1] / [(1＋sinθ)*cosθ]=0

2007-03-25 03:25:56 補充：
(cosθ)/(1 ＋ sinθ) - (1-sinθ)/(cosθ)=[cos^2θ - (1＋sinθ)*(1-sinθ) ] / [(1 ＋ sinθ)*cosθ] ...通分母=[cos^2θ - (1-sin^2θ) ] / [(1 ＋ sinθ)*cosθ]= [cos^2θ＋sin^2θ -1] / [(1＋sinθ)*cosθ]=0 改番好了...