Q.The value of \(\cfrac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1}\) is – (a) \(\cfrac{1+cos\theta}{sin\theta}\) (b) \(\cfrac{1+sin\theta}{cos\theta}\) (c) \(\cfrac{1+tan\theta}{sec\theta}\) (d) \(\cfrac{1+sec\theta}{sin\theta}\)
Answer: B
\(\cfrac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1}\)
\(= \cfrac{\sec\theta + \tan\theta - (\sec^2\theta - \tan^2\theta)}{\tan\theta - \sec\theta + 1}\)
\(= \cfrac{(\sec\theta + \tan\theta)(1 - \sec\theta + \tan\theta)}{\tan\theta - \sec\theta + 1}\)
\(= \sec\theta + \tan\theta\)
\(= \cfrac{1}{\cos\theta} + \cfrac{\sin\theta}{\cos\theta}\)
\(= \cfrac{1 + \sin\theta}{\cos\theta}\)
Similar Questions