Question

Given that $ x = \frac{\sin^2 \theta}{\tan \theta - \sec \theta}, \quad y = \frac{\sec \theta + \tan \theta}{\sec^2 \theta}, \quad \text{find} \, \frac{y}{x} $

• A. \( \frac{\sec \theta + \tan \theta}{\sin^2 \theta} \)
• B. \( \frac{\tan \theta + 1}{\sin \theta} \)
• C. \( \frac{\tan \theta + \sec \theta}{\sin^2 \theta} \)
• D. \( \frac{\sec \theta + \sin \theta}{\tan \theta} \)

Hint:

When simplifying trigonometric expressions, use known identities such as \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \sec \theta = \frac{1}{\cos \theta} \) to make the algebra easier.

Answer 1

The Correct Option is C

We are given: \[ x = \frac{\sin^2 \theta}{\tan \theta - \sec \theta}, \quad y = \frac{\sec \theta + \tan \theta}{\sec^2 \theta} \] We are asked to find \( \frac{y}{x} \). First, substitute the expressions for \( x \) and \( y \) into \( \frac{y}{x} \): \[ \frac{y}{x} = \frac{\frac{\sec \theta + \tan \theta}{\sec^2 \theta}}{\frac{\sin^2 \theta}{\tan \theta - \sec \theta}} \] Now simplify the expression by multiplying the numerator and the denominator: \[ \frac{y}{x} = \frac{\sec \theta + \tan \theta}{\sec^2 \theta} \times \frac{\tan \theta - \sec \theta}{\sin^2 \theta} \] Simplify the terms: \[ \frac{y}{x} = \frac{(\sec \theta + \tan \theta)(\tan \theta - \sec \theta)}{\sin^2 \theta \cdot \sec^2 \theta} \] Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we find the correct simplified expression to be: \[ \frac{\tan \theta + \sec \theta}{\sin^2 \theta} \] 
Thus, the correct answer is \( (C) \).