<p>The numerical value of \(\tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{3}\) is ______.</p>
Step-by-Step Solution
Key Concept: Use the addition formula for inverse tangent: tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1-ab)) when ab < 1. This directly converts the sum into a single inverse tangent that evaluates to a standard angle.
<p><strong>Step 1:</strong> Identify the formula for sum of inverse tangents.</p><p>When ab < 1: tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1-ab))</p><p><strong>Step 2:</strong> Verify the condition. Here a = 1/2, b = 1/3, so ab = 1/6 < 1 ✓</p><p><strong>Step 3:</strong> Apply the formula:</p><p>tan⁻¹(1/2) + tan⁻¹(1/3) = tan⁻¹((1/2 + 1/3)/(1 - 1/2·1/3))</p><p><strong>Step 4:</strong> Simplify the numerator: 1/2 + 1/3 = 3/6 + 2/6 = 5/6</p><p><strong>Step 5:</strong> Simplify the denominator: 1 - 1/6 = 5/6</p><p><strong>Step 6:</strong> Evaluate: tan⁻¹((5/6)/(5/6)) = tan⁻¹(1) = π/4</p><p>∴ Answer: <strong>π/4</strong></p>
Correct Answer: π/4