27)Prove the following identity. \(\sqrt{\frac{1+\sin\theta}{1-\sin\theta}} = \sec\theta + \tan\theta\)

27)Prove the following identity. \(\sqrt{\frac{1+\sin\theta}{1-\sin\theta}} = \sec\theta + \tan\theta\)

Answer:

LHS = \(\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}\)

Multiplying the Numerator and denominator by \(\sqrt{1+\sin\theta}\)

\(= \sqrt{\frac{1+\sin\theta}{1-\sin\theta} \times \frac{1+\sin\theta}{1+\sin\theta}} = \sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}\)

\(= \sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}}\)

\(= \frac{1+\sin\theta}{\cos\theta} = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\)

\(= \sec\theta + \tan\theta = \text{RHS}\)