Prove that $\displaystyle \frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$

\[\begin{aligned} & \text { L.H.S }=\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1} \\ & \text { Divide Numerator and Denominator by } \cos \theta \\ &=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\ &=\frac{\tan \theta-1+\sec \theta}{(\tan \theta-\sec \theta)+\left(\sec ^{2} \theta-\tan ^{2} \theta\right)} \\ &=\frac{\tan \theta-1+\sec \theta}{(\sec \theta-\tan \theta)(\tan \theta+\sec \theta-1)} \\ &=\frac{1}{\sec \theta-\tan \theta}=\text { R.H.S } \end{aligned} \]