The value of cot ( Σₙ₌₁⁵⁰ tan ⁻¹( 1/(1 + n + n²) ) ) is :

Solution

Maths sir · 4 steps

Write 1 + n + n² as 1 + n(n+1), and each term splits into a difference that telescopes.

1. Spot the pattern. The numerator is 1 = (n+1) - n and the denominator is 1 + n(n+1). That is exactly the shape of (a-b)/(1+ab) with a = n+1 and b = n.

   tan⁻¹(1/(1+n+n²)) = tan⁻¹(((n+1)-n)/(1+n(n+1))) = tan⁻¹(n+1) - tan⁻¹ n

2. Telescope the sum. Adding from n = 1 to 50, every middle term cancels: (tan⁻¹ 2 - tan⁻¹ 1) + (tan⁻¹ 3 - tan⁻¹ 2) + … + (tan⁻¹ 51 - tan⁻¹ 50). Only tan⁻¹ 51 - tan⁻¹ 1 is left.
3. Combine into one. tan⁻¹ 51 - tan⁻¹ 1 = tan⁻¹((51-1)/(1+51)) = tan⁻¹ 50/52 = tan⁻¹ 25/26.
4. So. The angle has tan = 25/26, so its cotangent is 26/25. The answer is (a).

The idea to keep. When an inverse tan term has 1 on top and 1 + (product of two consecutive things) below, split it as tan⁻¹ a - tan⁻¹ b and let the sum telescope.

Answer from the NTA final answer key. Solution by PrepWiser.