Question

Define a relation \( R \) on the interval \( \left[ 0, \frac{\pi}{2} \right] \) by \( x \, R \, y \) if and only if \( \sec^2 x - \tan^2 y = 1 \). Then \( R \) is:

• A. both reflexive and transitive but not symmetric
• B. both reflexive and symmetric but not transitive
• C. reflexive but neither symmetric nor transitive
• D. an equivalence relation

Hint:

To determine if a relation is an equivalence relation, check if it satisfies reflexivity, symmetry, and transitivity. If all three properties hold, it is an equivalence relation.

Answer 1

The Correct Option is D

Step 1: Recall the properties of equivalence relations: a relation \( R \) is an equivalence relation if it is reflexive, symmetric, and transitive. 
Step 2: - Reflexive: For every \( x \) in the given interval, \( x R x \) must hold. That is, we check if \( \sec^2 x - \tan^2 x = 1 \). This is true for all \( x \) in the interval \( \left[ 0, \frac{\pi}{2} \right] \), so the relation is reflexive. 
- Symmetric: For the relation to be symmetric, if \( x R y \), then \( y R x \) must also hold. Since the equation involves both \( x \) and \( y \) in a symmetric manner, the relation is symmetric. 
- Transitive: For transitivity, if \( x R y \) and \( y R z \), then \( x R z \) must hold. This property holds as well, meaning the relation is transitive. Thus, \( R \) is reflexive, symmetric, and transitive, so it is an equivalence relation.