JEE Advanced 2013 Paper 1Question 1
The value of $\cot\left(\sum_{n=1}^{23}\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right)\right)$ is
Inverse Trigonometric Functions
Solution
$1+\sum_{k=1}^n2k=n^2+n+1$ and $\cot^{-1}(n^2+n+1)=\tan^{-1}(n+1)-\tan^{-1}n$. The sum telescopes to $\tan^{-1}24-\tan^{-1}1$, whose cotangent is $\frac{1+24}{24-1}=\frac{25}{23}$.