中三數學問題求幫忙

1.(tan^2θ/sin^2θ)*2cosθ
(sin^2θ/cos^2θ)* (1/sin^2θ)*2cosθ
(1/cos^2θ)* 2cosθ
2/cos θ
2 sec θ

2.5tanθcosθ-4sinθ
5( sinθ /cosθ) cosθ-4sinθ
5sinθ - 4sinθ
sin θ

3.cos^3θ+sin^2θ cosθ
cos^2θ cos θ +sin^2θ cosθ
cos θ (cos^2θ + sin^2θ)
cos θ (1)
cos θ

4.1+2sin^2θ-cos^2θ
1+2sin^2θ- ( 1 - sin^2θ)
1+2sin^2θ - 1 + sin^2θ
2sin^2θ + sin^2θ
3sin^2θ

5.(sinθ-cosθ)^2
sin^2θ – 2sinθcosθ + cos^2
sin^2θ+ cos^2 – 2sinθcosθ
1 – 2sinθcosθ (Answer)

If you have learnt double angles,
1 – 2sinθcosθ
1 – sin 2θ (Answer)

6.tanθsinθcosθ+cos^2θ
( sinθ/cosθ )sinθcosθ+cos^2θ
sin^2θ + cos^2θ
1

7.2sin(90-θ)/cos(90-θ)
2cosθ/sinθ
2/tan θ or 2 cot θ

8.(sinθcosθ/tan(90-θ))+cos^2θ
(sinθcosθ/sin(90-θ)/ cos (90- θ)+cos^2θ
(sinθcosθ cos(90-θ)/ sin (90- θ)+cos^2θ
(sinθcosθ sinθ/cosθ)+cos^2θ
(sinθsinθ)+cos^2θ
sin^2θ + cos^2θ
1

9. cos^2(90-θ)/1+sin^2θ-cos^2θ
sin^2θ+sin^2θ-cos^2θ
2sin^2θ - cos^2θ

There may be a mistake in question 9

If cos^2(90-θ)/(1+sin^2θ-cos^2θ)
sin^2θ/(1+sin^2θ-(1- sin^2θ)
sin^2θ/(1+sin^2θ-1+ sin^2θ
sin^2θ/2sin^2θ
1/2


2013-08-28 07:45:47 補充：
You have to know these formulae :
sin^2θ + cos^2θ = 1
tan θ = sinθ/cos θ
sin (90 - θ) = cos θ
cos (90 - θ) = sin θ
sin 2θ = 2 sinθ cos θ
sec θ = 1/ cos θ