Trigonometry & Inverse Trigonometry Question - None

Question

The value of $ \cot\!\left(\displaystyle\sum_{n=1}^{19}\cot^{-1}\!\left(1+\displaystyle\sum_{p=1}^{n}2p\right)\right) $ is:

Accepted Answer

Step 1: Simplify the inner sum: ∑_p=1ⁿ 2p = 2·(n(n+1)/2) = n(n+1)

Step 2: The argument becomes 1 + n(n+1) = n² + n + 1 = (n+1)n + 1

Step 3: Use the telescoping identity: cot⁻¹(n² + n + 1) = cot⁻¹(n) - cot⁻¹(n+1)

Step 4: Apply this for n = 1 to 19:

∑_n=1¹⁹ cot⁻¹(1 + n(n+1)) = ∑_n=1¹⁹ [cot⁻¹(n) - cot⁻¹(n+1)]

Step 5: This telescopes to: cot⁻¹(1) - cot⁻¹(20) = π/4 - cot⁻¹(20)

Step 6: Therefore: cot(π/4 - cot⁻¹(20))

Step 7: Using cot(A - B) = (cot A cot B + 1)/(cot B - cot A) with A = π/4 and B = cot⁻¹(20):

cot(π/4 - cot⁻¹(20)) = (1·20 + 1)/(20 - 1) = 21/19

∴ Answer: 21/19

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