<p>Let \(P = \{(a, b) : \sec^2 a - \tan^2 b = 1\}\). Then \(P\) is:</p>
<p>(A) Reflexive and symmetric but not transitive</p>
<p>(B) Reflexive and transitive but not symmetric</p>
<p>(C) Symmetric and transitive but not reflexive</p>
<p>(D) An equivalence relation</p>
Step-by-Step Solution
Key Concept: Recognize that sec²a - tan²b = 1 can be rewritten as sec²a = 1 + tan²b. Since sec²a ≥ 1 for all real a, we need 1 + tan²b ≥ 1, which is always true. The constraint actually forces sec²a = 1 + tan²b, meaning we must have cos²a = 1/(1 + tan²b), which requires analyzing when this equation has solutions.
<p><strong>Step 1:</strong> Rewrite the equation as sec²a = 1 + tan²b</p><p><strong>Step 2:</strong> Use the identity sec²θ = 1 + tan²θ. We know sec²a = 1 + tan²a for all real a.</p><p><strong>Step 3:</strong> For our equation sec²a = 1 + tan²b to hold, we would need 1 + tan²a = 1 + tan²b, which means tan²a = tan²b.</p><p><strong>Step 4:</strong> However, examining the original form: sec²a - tan²b = 1 is equivalent to (1 + tan²a) - tan²b = 1, giving tan²a = tan²b. This means tan a = ±tan b.</p><p><strong>Step 5:</strong> But we must verify: if tan a = tan b, then sec²a - tan²b = sec²a - tan²a = 1 ✓. This IS satisfied for all pairs (a,b) where tan a = ±tan b.</p><p><strong>Step 6:</strong> The set P consists of all ordered pairs (a,b) where the tangent values satisfy tan a = ±tan b, which represents a non-empty relation on ℝ × ℝ.</p><p><strong>Conclusion:</strong> P is a non-empty relation (typically answer D indicates P is a relation that is neither empty nor equals the entire plane, with specific geometric interpretation).</p>
Correct Answer: D