關於數學三角學問題?

1.Cos(270°+θ)/Sin30°－Sin(180°－θ)/Tan135
=Sinθ/(1/2)－Sinθ/(－1)
=3Sinθ

2.(a) For 0°≤ θ≤ 360°，find the roots of equation sinx tanx=sinx
Sol
SinxTanx=Sinx
Sinx(Tanx－1)=0
Sinx=0 or Ranx=1
(1) Sinx=0
x=0° or x=180° or x=360°
(2) Tanx=1
x=45° or x=225°
(b) Find 1/Sin θ,Tan θ, given Cosθ=1/3
Cosθ=1/3
Sinθ=+/-2√2/3
1/Sinθ=+/-3√2/3
Tanθ=+/-2√2

1.
cos(270°＋ θ)/sin30° －sin(180°－ θ)/tan135
=(sinθ)/0.5 +(sinθ)/1
=3sinθ

2a
sinx tanx=sinx
(sin x)(tan x)-sin x=0
(sin x)(tan x -1)=0
sin x=0 or tan x=1
x=0 or x=180 or x=360 or x=45 or 225

2b
sin^2 θ+cos^θ=1
sin^2 θ=8/9
sinθ=(8/9)^0.5 or sinθ=-(8/9)^0.5
1/sinθ=3/(8)^0.5 or 1/sinθ=-3/(8)^0.5

cosθ=1/3
θ=70.52877937
tanθ
=tan70.52877937
=2.828427125

1.Cos(270°+θ)/Sin30°－Sin(180°－θ)/Tan135
=Sinθ/(1/2)－Sinθ/(－1)
=3Sinθ

2.(a) For 0°≤ θ≤ 360°，find the roots of equation sinx tanx=sinx
Sol
SinxTanx=Sinx
Sinx(Tanx－1)=0
Sinx=0 or Ranx=1
(1) Sinx=0
x=0° or x=180° or x=360°
(2) Tanx=1
x=45° or x=225°
(b) Find 1/Sin θ,Tan θ, given Cosθ=1/3
Cosθ=1/3
Sinθ=+/-2√2/3
1/Sinθ=+/-3√2/3
Tanθ=+/-2√2