三角比的關係

( sinθ - 3cosθ ) / ( cosθ - 3sinθ ) = 9/5
5(sinθ -3cosθ )=9(cosθ -3sinθ )
5sinθ -15cosθ =9cosθ -27sinθ
32sinθ =24cosθ
24/32=sinθ /cosθ
tanθ =3/4//

b) 從而，求3sinθ + 4cosθ
tanθ =3/4
so sinθ =3, cosθ =4
3sinθ +4cosθ =3x3+4x4=25//

sin(θ - 10) = cos 64
sin(θ-10)=sin(16)
θ-10=16
θ=26//

tanθ tan(θ+12) = 1
tanθ (tanθ +tan12)/(1-tanθ tan12)=1
tan^2 θ +tanθ tan12=1-tan θ tan12
tan^2 θ +2tanθ tan12 -1=0
......................................to be continue

a.( sinθ - 3cosθ ) / ( cosθ - 3sinθ ) = 9/5
5sinθ-15cosθ=9cosθ-27sinθ
32sinθ=24cosθ
4sinθ=3cosθ
sinθ/cosθ= 3/4
tanθ=3/4

b.考慮一個直角三角形，tan θ =對邊/鄰邊，
假設對邊是3, 鄰邊是4, 斜邊=√( 3^2+4^2)=5
cos θ= 4/5, sinθ=3/5
3sinθ + 4cosθ
=3*3/5 +4*4/5
=9/5+16/5
=5

sin(θ - 10°) = cos 64°
sin(θ - 10°) = sin(90°-64°)
sin(θ - 10°) = sin 26°
As you haven't give the range of θ, I suppose that you want to find the general solution, then
θ - 10°=180°n +(-1)^n 26°, where n is integer
θ=180°n +(-1)^n 26°+10°

tanθ tan(θ+12°) = 1
tan(θ+12°) = 1/tanθ
tan(θ+12°) = tan (90°-θ)
θ+12°=180°n+90°-θ, where n is integer
2θ=180°n+78°
θ=90°n+39°