Answer: B
\(\cfrac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1}\)
\(= \cfrac{\sec\theta + \tan\theta - (\sec^2\theta - \tan^2\theta)}{\tan\theta - \sec\theta + 1}\)
\(= \cfrac{(\sec\theta + \tan\theta)(1 - \sec\theta + \tan\theta)}{\tan\theta - \sec\theta + 1}\)
\(= \sec\theta + \tan\theta\)
\(= \cfrac{1}{\cos\theta} + \cfrac{\sin\theta}{\cos\theta}\)
\(= \cfrac{1 + \sin\theta}{\cos\theta}\)
\(\cfrac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1}\)
\(= \cfrac{\sec\theta + \tan\theta - (\sec^2\theta - \tan^2\theta)}{\tan\theta - \sec\theta + 1}\)
\(= \cfrac{(\sec\theta + \tan\theta)(1 - \sec\theta + \tan\theta)}{\tan\theta - \sec\theta + 1}\)
\(= \sec\theta + \tan\theta\)
\(= \cfrac{1}{\cos\theta} + \cfrac{\sin\theta}{\cos\theta}\)
\(= \cfrac{1 + \sin\theta}{\cos\theta}\)