Verified answer

a) The Identity is : Cos(A + B) = CosAcosB - Sin ASinB

Cosθ = Cos(½θ + ½θ) = Cos½θCos½θ - Sin½θSin½θ = Cos²(½θ) - Sin²(½θ)

Now use the Identity Cos²A + Sin²A = 1 ⇒⇒ Cos² = 1 - Sin²A

Cos²(½θ) - Sin²(½θ) = 1 - Sin²(½θ) - Sin²(½θ) = 1 - 2Sin²(½θ).

b) Use the Identity Sin(A + B) = SinACosB + CosASinB and the above proof.

Sinθ = Sin½θCos½θ + Cos½θSin½θ = 2Sin½θCos½θ

Then, 1 + Sinθ - Cosθ = 1 + 2Sin½θCos½θ - [1 - 2Sin²(½θ)] = 2Sin½θCos½θ + 2Sin²(½θ)

= 2Sin(½θ)[Cos(½θ) + Sin(½θ)]

NOTE : This isn't quite the answer provided by you but I'm hoping there's a typing error in your question and the coefficient of 2 has been overlooked.

c) If, 1 + Sinθ - Cosθ = 0, then we can substitute the alternative identity :-

2Sin(½θ)[Cos(½θ) + Sin(½θ)] = 0

So either Sin(½θ) = 0 then (½θ) = 0

or, Cos(½θ) + Sin(½θ) = 0 which means, Cos(½θ) = - Sin(½θ) : 1 = -Sin(½θ) / Cos(½θ = -Tan(½θ)

Tan(½θ) = -1 .......thus ½θ = ¾π and therefore θ = 3π/2 which is in the range 0 ≤ θ ≤ 2π.