Question

Define a relation \( R \) on the interval \( [0, \frac{\pi}{2}] \) by \( xRy \) 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 verify if a relation is an equivalence relation, check if it is reflexive, symmetric, and transitive.

Answer 1

The Correct Option is D

To determine if the relation \( R \) on the interval \( [0, \frac{\pi}{2}] \) given by \( xRy \) if and only if \( \sec^2 x - \tan^2 y = 1 \) is an equivalence relation, we must check if it satisfies the properties of reflexivity, symmetry, and transitivity.

• Reflexivity: A relation \( R \) is reflexive if every element is related to itself. Here, for \( xRx \), we need \( \sec^2 x - \tan^2 x = 1 \). We use the trigonometric identity \( \sec^2 x = 1 + \tan^2 x \), which simplifies to \( \sec^2 x - \tan^2 x = 1 \). Therefore, \( R \) is reflexive.
• Symmetry: A relation \( R \) is symmetric if for any elements \( x \) and \( y \), whenever \( xRy \), then \( yRx \). Here, if \( xRy \Rightarrow \sec^2 x - \tan^2 y = 1 \), then for \( yRx \), we need \( \sec^2 y - \tan^2 x = 1 \). Since \( \sec^2 y = 1 + \tan^2 y \) and \( \tan^2 x = \sec^2 x - 1 \), we have \( \sec^2 y - \tan^2 x = 1 + \tan^2 y - (\sec^2 x - 1) = 1 + \tan^2 y - \tan^2 y = 1 \), thus symmetry holds.
• Transitivity: A relation \( R \) is transitive if whenever \( xRy \) and \( yRz \), then \( xRz \). Assuming \( \sec^2 x - \tan^2 y = 1 \) and \( \sec^2 y - \tan^2 z = 1 \), then \( \sec^2 x - \tan^2 z = (\sec^2 x - \tan^2 y) + (\sec^2 y - \tan^2 z) - \sec^2 y \). Therefore, \( \sec^2 x - \tan^2 z = 1 + 1 - \sec^2 y \). Simplifying gives \( 2 - (1 + \tan^2 y) = 1 \), which holds as \( \sec^2 x - \tan^2 z = 1 \). Hence, transitivity is satisfied.

Since \( R \) satisfies reflexivity, symmetry, and transitivity, \( R \) is indeed an equivalence relation.