<p>In the following, each question has multiple answers. Choose the correct answer(s).</p><p>(a) If in a △ABC, the angle C is obtuse then tan A · tan B is</p>
Step-by-Step Solution
Key Concept: In a triangle, A + B + C = π, so A + B = π - C. When C is obtuse (C > π/2), we have A + B < π/2, making both A and B acute. Taking tangent of A + B = π - C and using tan(π - C) = -tan C reveals the relationship between tan A · tan B and (tan A + tan B).
<p><strong>Step 1:</strong> In triangle ABC, A + B + C = π, so A + B = π - C</p><p><strong>Step 2:</strong> Taking tangent: tan(A + B) = tan(π - C) = -tan C</p><p><strong>Step 3:</strong> Using tangent addition formula: tan(A + B) = (tan A + tan B)/(1 - tan A · tan B)</p><p><strong>Step 4:</strong> Therefore: (tan A + tan B)/(1 - tan A · tan B) = -tan C</p><p><strong>Step 5:</strong> Since C is obtuse, C ∈ (π/2, π), so A, B are both acute and A + B < π/2, meaning tan A > 0, tan B > 0</p><p><strong>Step 6:</strong> From Step 4: tan A + tan B = -tan C(1 - tan A · tan B). Since tan C < 0 (obtuse angle), the RHS is positive, which is consistent.</p><p><strong>Step 7:</strong> Rearranging: 1 - tan A · tan B = (tan A + tan B)/(-tan C). Since LHS must be positive (as tan A · tan B > 0 for acute angles), we get: <strong>tan A · tan B < 1</strong></p><p>∴ Answer: B</p>
Correct Answer: B