Question

The value of $\cot \left(\sum _{n=1}^{23}{\cot }^{-1}(1+\sum _{k=1}^{n}2k)\right)$ is

• A. $\frac{23}{25}$
• B. $\frac{25}{23}$
• C. $\frac{23}{24}$
• D. $\frac{24}{23}$

Answer 1

The correct option is B $\frac{25}{23}$

$\sum _{n=1}^{23}{\cot }^{-1}(1+\sum _{k=1}^{n}2k)$
$=\sum _{n=1}^{23}{\cot }^{-1}(1+n(n+1\left)\right)$
$=\sum _{n=1}^{23}{\tan }^{-1}\left(\frac{1}{1+n(n+1)}\right)$
$=\sum _{n=1}^{23}({\tan }^{-1}(n+1)-{\tan }^{-1}\left(n\right))$
$={\tan }^{-1}\left(24\right)-{\tan }^{-1}\left(1\right)$
$={\tan }^{-1}\left(\frac{24-1}{1+24}\right)$
$={\tan }^{-1}\left(\frac{23}{25}\right)$

Now, $\cot \left({\tan }^{-1}\left(\frac{23}{25}\right)\right)$

$=\cot \left({\cot }^{-1}\left(\frac{25}{23}\right)\right)$

$=\frac{25}{23}$