$$\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}$$ = ?

As per the given question,

$$\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}$$

Now taking the conjugate of the  given equation,

$$\Rightarrow \frac{(\sin \theta - \cos \theta + 1)(\sin \theta + \cos \theta + 1)}{(\sin \theta + \cos \theta - 1)(\sin \theta + \cos \theta + 1)}$$

$$\Rightarrow \dfrac{\sin^2 \theta + \sin\theta \cos \theta +\sin \theta- \sin \theta \cos \theta-\cos^2\theta-\cos\theta+\sin\theta+\cos\theta+1}{\sin^2 \theta + \sin\theta \cos \theta +\sin \theta+ \sin \theta \cos \theta+\cos^2\theta+\cos\theta0-\sin\theta-\cos\theta-1}$$

Here few terms are in equal and opposite sign so they will cancel out to each other

$$\Rightarrow \dfrac{\sin^2\theta+2\sin\theta}{2\sin\theta\cos\theta}$$

$$\Rightarrow \dfrac{\sin\theta}{\cos\theta}+\dfrac{1}{\cos\theta}$$

$$\Rightarrow \tan\theta+\sec\theta$$