有條數五識??help me

(1-cosθ)(1+cosθ)/cos^2 (90 °-θ)

= 1-cos^2 θ /cos^2 (90 ° - θ) [ Please remember that cos(90 °-θ)=sinθ ]

=sin^2 θ / sin^2 θ

=1

∵ L.H.S = R.H.S

∴(1-cos θ)(1+cos θ)/cos^2(90 ° - θ) ≡1

2007-02-07 19:44:06 補充：
加多點解釋 a, ^2 代表2次方b. (1-cosθ)(1 cosθ)= (1^2-cos^2θ)= 1-cos^2θc. cos^2 θ sin^2 θ=1