Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>Considering only the principal values of inverse functions, the set \( A = \left\{x \geq 0;\, \tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}\right\} \)</p>
<p>contains two elements</p>
<p>contains more than two elements</p>
<p>is a singleton</p>
<p>is an empty set</p>
Step-by-Step Solution
Key Concept: Use the tangent addition formula: tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1-ab)) when ab < 1, then solve the resulting equation for x using the principal value constraint.
<p><strong>Step 1:</strong> Apply the tangent addition formula. Since tan⁻¹(2x) + tan⁻¹(3x) = π/4, taking tangent of both sides:</p><p>tan[tan⁻¹(2x) + tan⁻¹(3x)] = tan(π/4) = 1</p><p><strong>Step 2:</strong> Use tan(A + B) = (tan A + tan B)/(1 - tan A tan B):</p><p>(2x + 3x)/(1 - 2x·3x) = 1</p><p>5x/(1 - 6x²) = 1</p><p><strong>Step 3:</strong> Solve for x:</p><p>5x = 1 - 6x²</p><p>6x² + 5x - 1 = 0</p><p>(6x - 1)(x + 1) = 0</p><p>x = 1/6 or x = -1</p><p><strong>Step 4:</strong> Apply constraints. Since x ≥ 0, reject x = -1. Verify x = 1/6 satisfies ab < 1: (2·1/6)(3·1/6) = 1/6 < 1 ✓</p><p><strong>Step 5:</strong> Verify solution: tan⁻¹(1/3) + tan⁻¹(1/2) = π/4 (using the addition formula backwards confirms this)</p><p>∴ A = {1/6}</p>
Correct Answer: C