Prove that :

$\dfrac{\text{sin θ - cos θ + 1}}{\text{sin θ + cos θ - 1}} = \dfrac{\text{cos θ}}{\text{1 - sin θ}}.$

To prove:

$\dfrac{\text{sin θ - cos θ + 1}}{\text{sin θ + cos θ - 1}} = \dfrac{\text{cos θ}}{\text{1 - sin θ}}.$

Multiplying numerator and denominator of L.H.S. by sin θ + cos θ + 1

$\Rightarrow \dfrac{\text{sin θ - cos θ + 1}}{\text{sin θ + cos θ - 1}} \times \dfrac{\text{sin θ + cos θ + 1}}{\text{sin θ + cos θ + 1}} \\[1em] \Rightarrow \dfrac{\text{(sin θ + 1) - cos θ}}{\text{(sin θ + cos θ) - 1}} \times \dfrac{\text{(sin θ + 1) + cos θ}}{\text{(sin θ + cos θ)} + 1} \\[1em] \Rightarrow \dfrac{\text{(sin θ + 1)}^2 - \text{cos}^2 θ}{\text{(sin θ + cos θ)}^2 - 1^2} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ + 1 + \text{2 sin θ - cos}^2 θ}{\text{sin}^2 θ + \text{cos}^2 θ + \text{ 2 sin θ cos θ - 1}} ……(1)$

Since, sin² θ + cos² θ = 1

⇒ sin² θ = 1 - cos² θ

Substituting these values in equation (1) :

$\Rightarrow \dfrac{\text{1 - cos}^2 θ + \text{1 + 2 sin θ - cos}^2 θ}{1 + \text{2 sin θ cos θ - 1}} \\[1em] \Rightarrow \dfrac{\text{2 + 2 sin θ - 2 cos}^2 θ}{\text{2 sin θ cos θ}} \\[1em] \Rightarrow \dfrac{\text{2(1 + sin θ - cos}^2 θ)}{\text{2 sin θ cos θ}} \\[1em] \Rightarrow \dfrac{\text{1 - cos}^2 \text{θ + sin θ} }{\text{sin θ cos θ}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 θ + \text{sin θ}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \dfrac{\text{sin θ(sin θ + 1)}}{\text{sin θ cos θ}} \\[1em] \Rightarrow \dfrac{\text{sin θ + 1}}{\text{cos θ}}$

Multiplying numerator and denominator by (1 - sin θ), we get :

$\Rightarrow \dfrac{\text{1 + sin θ}}{\text{cos θ}} \times \dfrac{\text{1 - sin θ}}{\text{1 - sin θ}} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 θ}{\text{cos θ(1 - sin θ)}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ}{\text{cos θ(1 - sin θ)}} \\[1em] \Rightarrow \dfrac{\text{cos θ}}{\text{1 - sin θ}}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin θ - cos θ + 1}}{\text{sin θ + cos θ - 1}} = \dfrac{\text{cos θ}}{\text{1 - sin θ}}.$