Question

If \( \sec \theta + \tan \theta = p \), then \( \frac{\sin \theta - 1}{\sin \theta + 1} \) is equal to:

• A. \( -\frac{1}{p^2} \)
• B. \( \frac{2}{p^2} \)
• C. \( \frac{1}{2p} \)
• D. \( -p^2 \)

Hint:

When given expressions involving \( \sec \theta \) and \( \tan \theta \), square the given equation to apply known trigonometric identities and simplify.

Answer 1

The Correct Option is A

We are given that \( \sec \theta + \tan \theta = p \). We need to find the value of: \[ \frac{\sin \theta - 1}{\sin \theta + 1} \] We can square both sides of the given equation \( \sec \theta + \tan \theta = p \): \[ (\sec \theta + \tan \theta)^2 = p^2 \] This expands to: \[ \sec^2 \theta + 2 \sec \theta \tan \theta + \tan^2 \theta = p^2 \] Using the identity \( \sec^2 \theta - \tan^2 \theta = 1 \), we substitute: \[ 1 + 2 \sec \theta \tan \theta = p^2 \] Now, express \( \sec \theta \tan \theta \) in terms of \( p \) and use this to simplify the given expression. Thus, the correct answer is \( -\frac{1}{p^2} \).