中三三角學問題

cosθ/ ( 1 + sinθ) - ( 1 - sinθ) / cosθ
= cos2θ/(1+sinθ)cosθ-(1-sinθ)(1+ sinθ)/cosθ(1+sinθ)
= [ cos2θ-(1-sinθ)(1+ sinθ)] / cosθ(1+sinθ)
= ( cos2θ- 1+ sin2θ) / cosθ(1+sinθ)
= ( 1 - 1 ) / cosθ(1+sinθ)
= 0


2008-04-02 22:48:18 補充：
其實只要通分同運用: sin^2 x + cos^2 x = 1就可以計到啦~~~

2008-04-02 22:57:08 補充：
即係第一步就通分, 令分母變成cosθ(1+sinθ);
第二、三步就化簡, 計到( cos^2θ- 1+ sin^2θ);
第四步就用sin^2 x + cos^2 x = 1將分子變成( 1 - 1 );
最後就會計到0。

cosθ/1+sinθ - 1-sinθ/cosθ

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

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

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

= 0