Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$ We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$
I have tried and it gets confusing.

multiply numerator and denominator by $(1-\sin\theta)$.
$$\dfrac {\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}\times \dfrac{1-\sin\theta}{1-\sin\theta}$$
$$\dfrac {\sin\theta-\sin^2\theta-\cos\theta+\sin\theta\cdot\cos\theta+1-\sin\theta}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$
$$\dfrac {1-\sin^2\theta-\cos\theta+\sin\theta\cdot\cos\theta}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$
$$\dfrac {\cos^2\theta-\cos\theta+\sin\theta\cdot\cos\theta}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$
$$\dfrac {\cos\theta(\cos\theta+\sin\theta-1)}{(\sin\theta+\cos\theta-1)(1-\sin\theta)}$$
$$\dfrac {\cos\theta}{1-\sin\theta}$$
$$\dfrac {1}{\dfrac{1-\sin\theta}{\cos\theta}}$$
$$\dfrac {1}{\dfrac{1}{\cos\theta}-\dfrac{\sin\theta}{{\cos\theta}}}$$
$$\dfrac {1}{\sec\theta-\tan\theta}$$
