<p>The numerical value of \(2\tan^{-1}\dfrac{1}{3} + \tan^{-1}\dfrac{1}{7}\) is ______.</p>
Step-by-Step Solution
Key Concept: Use the double angle formula for inverse tangent: 2tan⁻¹(a) = tan⁻¹(2a/(1-a²)), then apply the addition formula tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1-xy)) when xy < 1.
<p><strong>Step 1:</strong> Apply double angle formula to 2tan⁻¹(1/3):</p><p>2tan⁻¹(1/3) = tan⁻¹(2·(1/3)/(1-(1/3)²)) = tan⁻¹((2/3)/(1-1/9)) = tan⁻¹((2/3)/(8/9)) = tan⁻¹(6/8) = tan⁻¹(3/4)</p><p><strong>Step 2:</strong> Now add: tan⁻¹(3/4) + tan⁻¹(1/7)</p><p>Using tan⁻¹(x) + tan⁻¹(y) = tan⁻¹((x+y)/(1-xy)), where x = 3/4, y = 1/7:</p><p>xy = (3/4)(1/7) = 3/28 < 1 ✓</p><p><strong>Step 3:</strong> Calculate:</p><p>(x+y)/(1-xy) = (3/4 + 1/7)/(1 - 3/28) = ((21+4)/28)/(25/28) = (25/28)·(28/25) = 1</p><p><strong>Step 4:</strong> Therefore:</p><p>tan⁻¹(1) = π/4</p><p>∴ Answer: <strong>π/4</strong></p>
Correct Answer: π/4