Help Me With This Trig ID proof Please Soon!

Hello, Melcarthus!

Quote:

$\text{Prove: }\:\frac{\sin\theta -\cos\theta + 1}{\sin\theta + \cos\theta - 1} \:=\:\frac{\sin\theta + 1}{\cos\theta}$

The left side is: . $\frac{\sin\theta - (\cos\theta - 1)}{\sin\theta + (\cos\theta - 1)}$

Then: . $\frac{\sin\theta - (\cos\theta-1)}{\sin\theta + (\cos\theta - 1)} \cdot \color{blue}{\frac{\sin\theta - (\cos\theta - 1)}{\sin\theta - (\cos\theta -1)}}$

. . . . . $=\;\frac{\sin^2\!\theta - 2\sin\theta(\cos\theta-1) + (\cos\theta-1)^2}{\sin^2\!\theta - (\cos\theta -1)^2}$

. . . . . $=\;\frac{\sin^2\!\theta - 2\sin\theta\cos\theta + 2\sin\theta + \cos^2\!\theta - 2\cos\tjeta + 1}{\sin^2\!\theta - \cos^2\!\theta + 2\cos\theta - 1}$

. . . . . $=\;\frac{\overbrace{\sin^2\!\theta + \cos^2\!\theta}^{\text{This is 1}} + 1 + 2\sin\theta - 2\cos\theta - 2\sin\theta\cos\theta}{2\cos\theta - \cos^2\theta - \underbrace{(1-\sin^2\!\theta)}_{\text{This is }\cos^2\theta}}$

. . . . . $=\;\frac{2 + 2\sin\theta - 2\cos\theta - 2\sin\theta\cos\theta}{2\cos\theta - 2\cos^2\!\theta}$

. . . . . $=\;\frac{2(1+\sin\theta) - 2\cos\theta(1 + \sin\theta)}{2\cos\theta(1-\cos\theta)}$

. . . . . $=\;\frac{2(1+\sin\theta)(1-\cos\theta)}{2\cos\theta(1-\cos\theta)}$

. . . . . $=\;\frac{1+\sin\theta}{\cos\theta}$