<p>Let <em>P</em> be the relation defined on the set of all real numbers such that \(P = \{(a, b): \sec^2 a - \tan^2 b = 1\}\). Then <em>P</em> is</p>
<p>reflexive and symmetric but not transitive.</p>
<p>reflexive and transitive but not symmetric.</p>
<p>symmetric and transitive but not reflexive.</p>
<p>an equivalence relation.</p>
Step-by-Step Solution
Key Concept: Use the identity sec²a - tan²b = 1 to determine constraints on a and b. Recognize that sec²a ≥ 1 for all real a, so tan²b ≥ 0 is automatically satisfied, but we need sec²a = 1 + tan²b. This means for every a, there exist valid b values, and for every b, there exist valid a values, making P both reflexive-like in structure and symmetric.
<p><strong>Step 1:</strong> Analyze the condition sec²a - tan²b = 1, which gives us sec²a = 1 + tan²b.</p><p><strong>Step 2:</strong> Since sec²a ≥ 1 for all real a (with a ≠ π/2 + nπ), we need tan²b ≥ 0, which is always true for real b.</p><p><strong>Step 3:</strong> For any real a, sec²a takes values in [1, ∞). For tan²b to equal sec²a - 1, we need tan²b ∈ [0, ∞), which is satisfied by all real b.</p><p><strong>Step 4:</strong> Check symmetry: If (a,b) ∈ P, then sec²a - tan²b = 1. For (b,a): we'd need sec²b - tan²a = 1, which is a different condition. So P is <strong>NOT symmetric</strong>.</p><p><strong>Step 5:</strong> Check reflexivity: (a,a) ∈ P requires sec²a - tan²a = 1, which is TRUE for all a in domain. So P is <strong>reflexive</strong>.</p><p><strong>Step 6:</strong> Check transitivity: If (a,b) and (b,c) ∈ P, we have sec²a = 1 + tan²b and sec²b = 1 + tan²c. This does NOT guarantee sec²a - tan²c = 1. So P is <strong>NOT transitive</strong>.</p><p>∴ Answer: <strong>D</strong> (P is reflexive but neither symmetric nor transitive)</p>
Correct Answer: D