hard f.3 maths question!thanks

(tanΘ/sinΘ - cosΘ) (1/sinΘ - sinΘ)
=(sinΘ/cosΘ × 1/sinΘ- cosΘ)(1/sinΘ - sinΘ)
=(1/cosΘ- cosΘ)(1/sinΘ - sinΘ)
=[1-cos^2Θ)/cosΘ][(1-sin^2Θ)/ sinΘ]
=(sin^2Θ/cosΘ)(cos^2/ sinΘ)
=sinΘcosΘ



tan(90° - Θ) = 2, and Θ is an acute angle, find:
1)sinΘ,cosΘ
tan(90°-Θ)=2
1/tanΘ=2
tanΘ=1/2
☆對邊=1，鄰邊=2，斜邊=√(2^2+1^2)=√5
sinΘ=1/√5
cosΘ=2/√5

2) 2sinΘcosΘ/ 1+cos^2Θ-sin^2 Θ
=2sinΘcosΘ/ 1-sin^2+cos^2 Θ
=2sinΘcosΘ/cos^2 Θ+cos^2 Θ
=2sinΘcosΘ/2cos^2 Θ
=sinΘ/cosΘ
=tanΘ
3) Tan20°=1/Tan(Θ-24°)
Tan20°=Tan[90°-(Θ-24°)]
Tan20°=Tan(90°-Θ+24°)
Tan20°=Tan(114°-Θ)
20°=114°-Θ
Θ=94°








Tan20°=1/Tan(Θ-24°)

2010-05-30 18:50:34 補充：
2)如果你是計數值
answer會=1/2
你要將(1)sinΘ,cosΘ的值代回去