Question

If \( \sec(\theta + \alpha) \), \( \sec(\theta) \), and \( \sec(\theta - \alpha) \) are in arithmetic progression, then \( \sin^2 \theta = \)

• A. \( \cos \alpha \)
• B. \( 2 \cos \alpha \)
• C. \( -2 \cos \alpha \)
• D. \( -\cos \alpha \)

Hint:

In arithmetic progressions involving trigonometric functions, use the identities for sum and difference to simplify the terms.

Answer 1

The Correct Option is D

We are given that \( \sec(\theta + \alpha) \), \( \sec(\theta) \), and \( \sec(\theta - \alpha) \) are in arithmetic progression. In arithmetic progression, the middle term is the average of the other two terms: \[ \sec(\theta) = \frac{\sec(\theta + \alpha) + \sec(\theta - \alpha)}{2} \] Step 1: Use the sum and difference identities for secant. Using the identity for secant: \[ \sec(\theta + \alpha) + \sec(\theta - \alpha) = 2 \sec(\theta) \cos \alpha \] Thus, we can write: \[ \sec(\theta) = \sec(\theta) \cos \alpha \] Step 2: Solve for \( \sin^2 \theta \). Rearranging: \[ 1 = \cos \alpha \] So, \( \sin^2 \theta = 1 - \cos^2 \theta = - \cos \alpha \). Thus, the answer is \( -\cos \alpha \), which corresponds to option (4).