Prove that $\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity $\sec ^{2} \theta=1+\tan ^{2} \theta.$

(A) To prove the identity,we divide the numerator and the denominator of the $LHS$ by $\cos \theta$ to express it in terms of $\tan \theta$ and $\sec \theta.$
$LHS = \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta}$
Rearranging the terms,we get:
$LHS = \frac{(\tan \theta + \sec \theta) - 1}{(\tan \theta - \sec \theta) + 1}$
Using the identity $\sec^2 \theta - \tan^2 \theta = 1,$ we substitute $1$ in the numerator:
$LHS = \frac{(\tan \theta + \sec \theta) - (\sec^2 \theta - \tan^2 \theta)}{(\tan \theta - \sec \theta) + 1}$
Factorizing the numerator using $a^2 - b^2 = (a - b)(a + b)$:
$LHS = \frac{(\tan \theta + \sec \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{(\tan \theta - \sec \theta) + 1}$
Taking $(\tan \theta + \sec \theta)$ as a common factor:
$LHS = \frac{(\tan \theta + \sec \theta) [1 - (\sec \theta - \tan \theta)]}{(\tan \theta - \sec \theta) + 1}$
$LHS = \frac{(\tan \theta + \sec \theta) [1 - \sec \theta + \tan \theta]}{(\tan \theta - \sec \theta + 1)}$
Canceling the common term $(\tan \theta - \sec \theta + 1)$:
$LHS = \tan \theta + \sec \theta$
To get the $RHS,$ multiply and divide by $(\sec \theta - \tan \theta)$:
$LHS = \frac{(\sec \theta + \tan \theta)(\sec \theta - \tan \theta)}{\sec \theta - \tan \theta} = \frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta - \tan \theta} = \frac{1}{\sec \theta - \tan \theta} = RHS.$