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:
Show Hint
- both reflexive and transitive but not symmetric
- both reflexive and symmetric but not transitive
- reflexive but neither symmetric nor transitive
- an equivalence relation
The Correct Option is D
Solution and Explanation
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.
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