Trigonometry & Inverse Trigonometry
General
Grade 12

Question:

<p><span class="math-inline">\(\displaystyle\sum_{n=1}^{\infty}\tan^{-1}\!\frac{2n}{n^4+n^2+2}=\)</span></p>
π/2
π/4
π
π/3

Step-by-Step Solution

Key Concept: General
<div class="solution"><p><strong>Key Idea:</strong> Factor denominator as <span class="math-inline">$1+(n^2+n+1)(n^2-n+1)$</span>.</p><p><strong>Step 1:</strong> <span class="math-block">$$T_n=\tan^{-1}\!\frac{(n^2+n+1)-(n^2-n+1)}{1+(n^2+n+1)(n^2-n+1)}=\tan^{-1}(n^2+n+1)-\tan^{-1}(n^2-n+1)$$</span></p><p><strong>Step 2:</strong> Telescoping: <span class="math-inline">$\lim_{N\to\infty}[\tan^{-1}(N^2+N+1)-\tan^{-1}(1)]=\pi/2-\pi/4=\pi/4$</span>.</p><p><strong>Answer: (B) π/4</strong></p><div class="trap-box"><strong>Trap:</strong> Expanding the quartic blindly kills the telescope. Spot the product of two neighboring quadratics.</div><div class="key-concept"><strong>Key Concept:</strong> Quartic denominators hide product of neighboring quadratics — standard ITF telescope</div></div>
Correct Answer: 2

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