JEE Main 2013 Mathematics question on Inverse Trigonometric Functions

The inner sum ∑(k=1 to n) 2k = n(n+1), so each term becomes cot⁻¹(1 + n(n+1)) = cot⁻¹(n² + n + 1).

Now use the identity: cot⁻¹(n² + n + 1) = tan⁻¹(1/(n²+n+1)) = tan⁻¹((n+1) - n)/(1 + n(n+1)) = tan⁻¹(n+1) - tan⁻¹(n).

So the sum telescopes: ∑(n=1 to 23) [tan⁻¹(n+1) - tan⁻¹(n)] = tan⁻¹(24) - tan⁻¹(1).

Therefore cot(tan⁻¹(24) - tan⁻¹(1)) = cot(tan⁻¹((24-1)/(1+24))) = cot(tan⁻¹(23/25)) = 25/23.