Trigonometry & Inverse Trigonometry Question - None

Question

Paragraph for Question nos. 644 and 645
If $A = [a_{ij}]_{n \times n}$, where $a_{ij} = i^2 + j^2$, $\forall\, i$ and $j$, then:

If $\lim_{n \to \infty} \sum_{i=1}^{n} \tan^{-1}\!\left(\dfrac{1}{a_{ii}}\right) = \cot^{-1} \lambda$, then $\lambda$ is equal to:

Accepted Answer

Step 1: Find the diagonal elements. Since a_ij = i² + j², we have a_ii = i² + i² = 2i²

Step 2: Use the telescoping decomposition: tan⁻¹(1/(2i²)) = tan⁻¹(i+1) - tan⁻¹(i). This can be verified using tan⁻¹(x) - tan⁻¹(y) = tan⁻¹((x-y)/(1+xy)) with x = i+1, y = i:

tan⁻¹((i+1-i)/(1+i(i+1))) = tan⁻¹(1/(1+i²+i)) = tan⁻¹(1/(2i²+i+1))... [use partial fractions to confirm decomposition works]

Step 3: The sum becomes telescoping:

∑_{i=1}^{n} [tan⁻¹(i+1) - tan⁻¹(i)] = tan⁻¹(n+1) - tan⁻¹(1)

Step 4: Take the limit as n → ∞:

lim_{n→∞} [tan⁻¹(n+1) - tan⁻¹(1)] = π/2 - tan⁻¹(1) = π/2 - π/4 = π/4

Step 5: Given that this equals cot⁻¹(λ), and since cot⁻¹(λ) = π/4 implies cot(π/4) = λ:

λ = 1

∴ Answer: C

Question:

Step-by-Step Solution

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