Factoring trig functions quick question <_<?

Question

How did this:

(cosθ sinθ + (1 + sinθ) cosθ) / (cosθ cosθ - (1 + sinθ) sinθ)

become this:

(cosθ(1 + 2sinθ)) / (1 - 2sin^2θ - sinθ) ??

Answer 1

Numerator , N
cos θ (sin θ + 1 + sin θ)
cos θ (2 sin θ + 1)

Denominator, D
cos ² θ - sin θ - sin ² θ
cos ² θ - sin θ - (1 - cos ² θ)
2 cos ² θ - sin θ - 1
2 - 2 sin ² θ - sin θ - 1
1 - sin θ - 2 sin ² θ
(1 - 2 sin θ)(1 + sin θ)

N/D
(cos θ) (2 sinθ + 1) / (1 - sin θ)(1 + sin θ)

Answer 2

(cosθ sinθ + (1 + sinθ) cosθ) / (cosθ cosθ - (1 + sinθ) sinθ)
(cosθ sinθ + sinθcosθ + cosθ ) / (cosθ cosθ - sinθsinθ- sinθ)

numerator:
cosθ sinθ + sinθcosθ + cosθ =
2sinθcosθ + cosθ = cosθ (2sinθcosθ + 1)

denominator:
cosθ cosθ - sinθsinθ- sinθ =
cos^2 θ - sin^2 θ- sinθ =
(1 - sin^2 θ) - sin^2 θ- sinθ =
1- 2sin^2 θ - sinθ

There you have it.
Numerator/Denominator =
(cosθ(1 + 2sinθ)) / (1 - 2sin^2θ - sinθ)

Answer 3

let me use x instead of Θ for ease of typing.
top is cos x sin x + (1 + sin x)cos x. factor out cos x:
[sin x + (1 + sin x)]cos x and simplify to
[1 + 2sin x]cos x

bottom is [cos x cos x - (1 + sin x)sin x] =
[cos² x - sin x - sin² x] =
[1 - sin² x - sin x - sin² x] =
1 - 2sin² x - sin x

Answer 4

By using sin(2x) = 2 sin(x) cox(x).