Please answer the following question?

40.

(a) 2 cot(φ+35°) = 5

cot(φ+35°) = 5/2

tan(φ+35°) = 2/5

φ+35° = atan(2/5) ± k．180°

       = 21.8° ± k．180°

φ = -13.2° ± k．180°

∴ φ = 166.8° or 346.8°

(b)(i)

secθ/(cotθ+tanθ)

   = (1/cosθ)/(cosθ/sinθ + sinθ/cosθ)

   = 1/(cos^2(θ)/sinθ + sinθ)

   = 1/(cos^2(θ)+sin^2(θ))/sinθ)

   = 1/(1/sinθ)

   = sinθ

(ii) sec(3θ)/(cot(3θ)+tan(3θ)) = -√3/2

∴ sin(3θ) = -√3/2

-3π/2 ≦ 3θ ≦ 3π/2

∴ 3θ = -π/3 or 4π/3

∴ θ = -π/9 or 4π/9

41. cosecθ = 3sinθ+cotθ

∴ 1/sinθ = 3sinθ+cosθ/sinθ

0° < θ < 180°,

∴ sinθ ≠ 0,

方程式兩邊同乘以 sinθ, 得

   1 = 3sin^2(θ) + cosθ

     = 3 - 3cos^2(θ) + cosθ

∴ 3cos^2(θ) - cos(θ) - 2 = 0

∴ cosθ = [1 ± √(1+24)]/6 = 1 or -2/3

但 0° < θ < 180°, cosθ ≠ 1,

∴ cosθ = -2/3

   θ = acos(-2/3) = 131.81°

42.(i)

2cos(x)cot(x) + 1 = cot(x) + 2cos(x)

∴ 2cos(x)cot(x) - cot(x) - 2cos(x) + 1 = 0

∴ (2cos(x)-1)(cot(x)-1) = 0

cot(x) 有定義, 則 sin(x)≠0.

方程式兩邊乘以 sin(x), 則得等價方程式:

    (2cos(x)-1)(cos(x)-sin(x)) = 0

∴ a = 2, b = 1.

(ii)

∴ cos(x) = 1/2 or cos(x) = sin(x)

0 < x < π

    cos(x) = 1/2 if and only if x = π/3

    cos(x) = sin(x) if and only if x = π/4