Trigonometry & Inverse Trigonometry
Inverse Trigonometric Series
Grade 12
Question:
<p>If \( S_n = \sum_{r=1}^{n} \cot^{-1}(r^2 + 3r + 3) \), then:</p>
<p>(a) \( S_\infty = \cot^{-1}(2) \)</p>
<p>(b) \( S_5 = \cot^{-1}(3) \)</p>
<p>(c) \( S_6 = \cot^{-1}\left(\dfrac{17}{6}\right) \)</p>
<p>(d) \( S_8 = \cot^{-1}(5) \)</p>
Step-by-Step Solution
Key Concept: Use the telescoping property: cot⁻¹(r² + 3r + 3) = cot⁻¹((r+1)(r+2)) can be decomposed as tan⁻¹(r+1) - tan⁻¹(r) using the identity cot⁻¹(x) = tan⁻¹(1/x) and partial fractions.
<p><strong>Step 1:</strong> Recognize the denominator pattern: r² + 3r + 3 = r² + 3r + 2 + 1 = (r+1)(r+2) + 1</p><p><strong>Step 2:</strong> Apply the identity: cot⁻¹((r+1)(r+2) + 1) = tan⁻¹(r+2) - tan⁻¹(r+1) [using cot⁻¹(xy+1) = tan⁻¹(y) - tan⁻¹(x) where x = r+1, y = r+2]</p><p><strong>Step 3:</strong> Write the sum as a telescoping series:</p><p>Sₙ = [tan⁻¹(3) - tan⁻¹(2)] + [tan⁻¹(4) - tan⁻¹(3)] + [tan⁻¹(5) - tan⁻¹(4)] + ... + [tan⁻¹(n+2) - tan⁻¹(n+1)]</p><p><strong>Step 4:</strong> Most terms cancel, leaving: Sₙ = tan⁻¹(n+2) - tan⁻¹(2)</p><p><strong>Step 5:</strong> Apply tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab)) with a = n+2, b = 2:</p><p>Sₙ = tan⁻¹((n+2-2)/(1+2(n+2))) = tan⁻¹(n/(2n+5))</p><p>∴ Answer: A</p>
Correct Answer: A