Prove that

$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity

$\sec ^{2} \theta=1+\tan ^{2} \theta$

Since we will apply the identity involving $\sec \theta$ and $\tan \theta,$ let us first convert the $LHS$ (of the identity we need to prove) in terms of $\sec \theta$ and $\tan \theta$ by dividing numerator and denominator by $\cos \theta .$

$LHS=\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta}$

$=\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}=\frac{\{(\tan \theta+\sec \theta)-1\}(\tan \theta-\sec \theta)}{\{(\tan \theta-\sec \theta)+1\}(\tan \theta-\sec \theta)}$

$=\frac{\left(\tan ^{2} \theta-\sec ^{2} \theta\right)-(\tan \theta-\sec \theta)}{\{\tan \theta-\sec \theta+1\}(\tan \theta-\sec \theta)}$

$=\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)}$

$=\frac{-1}{\tan \theta-\sec \theta}=\frac{1}{\sec \theta-\tan \theta}$

which is the RHS of the identity, we are required to prove.