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

L.H.S. \(\displaystyle =\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}\) Dividing Numerator and denominator by \(\displaystyle \cos \theta\), we get \[\begin{aligned} & =\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\ & =\frac{(\sec \theta+\tan \theta)-\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{\tan \theta+1-\sec \theta} \\ & =\frac{(\sec \theta+\tan \theta)(1-\sec \theta+\tan \theta)}{\tan \theta+1-\sec \theta} \\ & =\sec \theta+\tan \theta \\ & =(\sec \theta+\tan \theta) \times \frac{(\sec \theta-\tan \theta)}{(\sec \theta-\tan \theta)} \\ & =\frac{\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{(\sec \theta-\tan \theta)} \\ & =\frac{1}{\sec \theta-\tan \theta}=\text { R.H.S. } \end{aligned} \]