中三數學題(兩題)

1)cos(90 ﹣x)[1 + tan2 (90 ﹣x)]
= (sin x){1 + [sin2 (90 ﹣x)/cos2 (90 ﹣x)]}
= (sin x){[cos2(90 ﹣x)/cos2(90 ﹣x)] + [sin2 (90 ﹣x)/cos2(90 ﹣x)]}
= (sin x){[cos2(90 ﹣x) + sin2 (90 ﹣x)]/cos2 (90 ﹣x)]}
= (sin x)[1/cos2 (90 ﹣x)]
= (sin x)(1/sin2 x)
= 1 / sinx

2)tan(90 ﹣θ) = 2/3
1 / tanθ= 2 / 3
tanθ= 3 / 2
∴sinθ= 3 / √13 = 3√13 / 13
cosθ= 2 / √13 = 2√13 / 13
∴(sinθcosθ)/ (1﹣sin2θ+ cos2θ )
= (3√13 / 13)(2√13 / 13)/ [1﹣(3√13 / 13)2+ (2√13 / 13)2]
= [78/ (13)(13)] / [1﹣(117/ 169)+ (52/ 169)]
= (78/ 169) / [1﹣(9/ 13)+ (4/ 13)]
= (6/ 13) / (8/ 13)
= 6 / 8
= 3 / 4

2008-04-20 17:24:09 補充：
You should use the Pythagorean theorem to find the values of sinθ and cosθ.

2008-04-20 17:25:31 補充：
tan(90﹣x) = 1/tanx
tanx = sinx/cosx
sin^2 x+ cos^2 x = 1
cos(90﹣x) = sinx
sin(90﹣x) = cosx

1. cos ( 90°- x ) [1+tan^2 ( 90°-x)]
=sin x (1+ 1/(tan^2 x)
=sin x (1+ cos^2 x / sin^2 x)
=sin x ((sin^2 x + cos^2 x) / sin^2 x)
=sin x (1 / sin^2 x)
=1 / sin x

2.tan ( 90°-θ) = 2/3
tan θ=3/2
sin θ : cos θ =3:2
sin^2 θ : cos^2 θ =9:4

sin^2 θ + cos^2 θ = 1

sin θ =root(1/(9+4)*9)
= 3(root13) / 13

cosθ=root(1-sin^2 θ)
= 2(root13) / 13

sinθcosθ/ (1-sin^2θ+ cos^2θ )
= (3(root13) / 13)*(2(root13) / 13) / (1-(9/13)+(4/13))
= (6/13) / (8/13)
= 3/4