(sin(2θ)+2cos(θ)-1)/(sin(2θ)+3cos(θ)-3)≡(cos(2θ)-cos(θ))/-sin(2θ) is not true as presented (see below) so we must look for other interpretations of the question (which is ambiguous); for example, do you mean sin(2θ) or sin2(θ)?
sin(2θ)=2sin(θ)cos(θ); cos(2θ)=1-2sin2(θ), so sin2(θ)=½(1-cos(2θ)).
When θ=π/2 the LHS=(0+0-1)/(0+0-3)=⅓, RHS is undefinable so LHS≠RHS invalidating the identity.
Let's next consider the interpretation:
(sin2(θ)+2cos(θ)-1)/(sin2(θ)+3cos(θ)-3)≡(cos2(θ)-cos(θ))/-sin2(θ).
When θ=π the LHS=(0-2-1)/(0-3-3)=½, RHS is undefinable so LHS≠RHS invalidating the identity.
However, a phase shift in the RHS makes LHS=RHS:
(sin2(θ)+2cos(θ)-1)/(sin2(θ)+3cos(θ)-3)≡(cos2(π-θ)-cos(π-θ))/-sin2(π-θ).
The RHS can be rewritten (cos2(θ)+cos(θ))/-sin2(θ).
Let's assume that the question was to prove the identity:
(sin2(θ)+2cos(θ)-1)/(sin2(θ)+3cos(θ)-3)≡(cos2(θ)+cos(θ))/-sin2(θ).
LHS:
(2cos(θ)-cos2(θ))/(-2-cos2(θ)+3cos(θ))=-cos(θ)(2-cosθ)/(cos2(θ)-3cos(θ)+2)=
-cos(θ)(2-cosθ)/(cos(θ)-2)(cos(θ)-1)=cos(θ)/(cos(θ)-1).
RHS:
(cos2(θ)+cos(θ))/-sin2(θ)=cos(θ)(cos(θ)+1)/(cos2(θ)-1)=
cos(θ)(cos(θ)+1)/((cos(θ)-1)(cos(θ)+1))=cos(θ)/(cos(θ)-1)=LHS, confirming the identity.