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 verify if the relation \( R \) defined on the interval \( [0, \frac{\pi}{2}] \) by \( xRy \) if and only if \( \sec^2 x - \tan^2 y = 1 \) is an equivalence relation, we must assess its reflexivity, symmetry, and transitivity.

• Reflexivity: A relation \( R \) is reflexive if \( xRx \) for all \( x \). For \( R \) to be reflexive, \( \sec^2 x - \tan^2 x \) must equal 1. The trigonometric identity \( \sec^2 x = 1 + \tan^2 x \) confirms that \( \sec^2 x - \tan^2 x = 1 \). Thus, \( R \) is reflexive.
• Symmetry: A relation \( R \) is symmetric if \( xRy \) implies \( yRx \) for all \( x, y \). Given \( xRy \Rightarrow \sec^2 x - \tan^2 y = 1 \). For \( yRx \), we require \( \sec^2 y - \tan^2 x = 1 \). Substituting \( \sec^2 y = 1 + \tan^2 y \) and \( \tan^2 x = \sec^2 x - 1 \), we get \( \sec^2 y - \tan^2 x = (1 + \tan^2 y) - (\sec^2 x - 1) = 1 + \tan^2 y - \sec^2 x + 1 = 2 + \tan^2 y - \sec^2 x \). From \( \sec^2 x - \tan^2 y = 1 \), we have \( \sec^2 x = 1 + \tan^2 y \). Substituting this, we obtain \( 2 + \tan^2 y - (1 + \tan^2 y) = 2 + \tan^2 y - 1 - \tan^2 y = 1 \). Therefore, symmetry holds.
• Transitivity: A relation \( R \) is transitive if \( xRy \) and \( yRz \) imply \( xRz \) for all \( x, y, z \). Assuming \( \sec^2 x - \tan^2 y = 1 \) and \( \sec^2 y - \tan^2 z = 1 \). From these, we have \( \sec^2 x = 1 + \tan^2 y \) and \( \sec^2 y = 1 + \tan^2 z \). Substituting the second into the first: \( \sec^2 x = 1 + (1 + \tan^2 z) = 2 + \tan^2 z \). We need to check if \( \sec^2 x - \tan^2 z = 1 \). Substituting our derived \( \sec^2 x \), we get \( (2 + \tan^2 z) - \tan^2 z = 2 eq 1 \). Re-evaluation of transitivity: From \( \sec^2 x - \tan^2 y = 1 \), \( \sec^2 x = 1 + \tan^2 y \). From \( \sec^2 y - \tan^2 z = 1 \), \( \tan^2 z = \sec^2 y - 1 \). We require \( \sec^2 x - \tan^2 z = 1 \). Substituting \( \tan^2 z \): \( \sec^2 x - (\sec^2 y - 1) = \sec^2 x - \sec^2 y + 1 \). This expression must equal 1. So, \( \sec^2 x - \sec^2 y = 0 \), meaning \( \sec^2 x = \sec^2 y \). This implies \( 1 + \tan^2 x = 1 + \tan^2 y \), so \( \tan^2 x = \tan^2 y \). This is not generally true given \( \sec^2 x - \tan^2 y = 1 \) and \( \sec^2 y - \tan^2 z = 1 \). Therefore, transitivity is not satisfied.

Since \( R \) does not satisfy transitivity, \( R \) is not an equivalence relation.