Trigonometry & Inverse Trigonometry
General
Grade 12
Question:
<p>\(\displaystyle\sum_{r=0}^{\infty}\tan^{-1}\!\frac{1}{r^2+3r+3}=\)</p>
\pi/2
<strong>\pi/4</strong>
cot⁻^13
tan⁻^12
Step-by-Step Solution
<div class="solution"><p><strong>Key Idea:</strong> Factor denominator: $r^2+3r+3=1+(r+1)(r+2)$.</p><p><strong>Step 1:</strong> <span class="math-block">$$T_r=\tan^{-1}\!\frac{(r+2)-(r+1)}{1+(r+2)(r+1)}=\tan^{-1}(r+2)-\tan^{-1}(r+1)$$</p><p><strong>Step 2:</strong> Telescoping sum: $\lim_{n\to\infty}[\tan^{-1}(n+2)-\tan^{-1}(1)]=\pi/2-\pi/4=\pi/4$.</p><p><strong>Answer: (B) $\pi/4$</strong></p><div class="trap-box"><strong>Trap:</strong> Miscounting the starting index -- the first term at r=0 must be verified.<div class="key-concept"><strong>Key Concept:</strong> tan⁻^1 telescoping sum -- factor denominator as 1+product of consecutive terms
Correct Answer: 2