Question

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

• A. \(2\sec\theta\)
• B. \(\sec\theta + \tan\theta\)
• C. \(\cos\theta - \cot\theta\)
• D. \(2\tan\theta\)

Answer 1

The Correct Option is A

Let’s consider the expression: \[ \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}} + \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} \] Multiply both terms by \( \sqrt{\frac{1 - \sin \theta}{1 - \sin \theta}} \) and \( \sqrt{\frac{1 + \sin \theta}{1 + \sin \theta}} \) respectively: \[ = \frac{1 + \sin \theta}{\sqrt{1 - \sin^2 \theta}} + \frac{1 - \sin \theta}{\sqrt{1 - \sin^2 \theta}} \] We know that: \[ 1 - \sin^2 \theta = \cos^2 \theta \Rightarrow \sqrt{1 - \sin^2 \theta} = \cos \theta \] So the expression becomes: \[ = \frac{1 + \sin \theta}{\cos \theta} + \frac{1 - \sin \theta}{\cos \theta} \] \[ = \frac{(1 + \sin \theta) + (1 - \sin \theta)}{\cos \theta} = \frac{2}{\cos \theta} = 2 \sec \theta \]

The correct answer is option (A): \(2\sec\theta\)