Inverse Trigonometric Functions
Summation Series
Premium Question
Grade 12
Question:
The value of $\cot\left[\sum_{n=1}^{23} \cot^{-1}\left(1 + \sum_{k=1}^n 2k\right)\right]$ is:
(1) $\frac{23}{25}$
(2) $\frac{25}{23}$
(3) $\frac{23}{24}$
(4) $\frac{24}{23}$
Step-by-Step Solution
Key Concept: Simplify $1+\sum_{k=1}^n 2k = n^2 +n+1$. Then show $\cot^{-1}(n^2 +n+1) = \tan^{-1}(n+1)-\tan^{-1}(n)$ and telescope.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)