Question

The value of $\cot \left(\sum _{n=1}^{50}{\tan }^{-1}\left(\frac{1}{1+n+n^2}\right)\right)$ is

• A. $\frac{26}{25}$
• B. $\frac{25}{26}$
• C. $\frac{50}{51}$
• D. $\frac{52}{51}$

Answer 1

Answer: (A)

${\tan }^{-1}\frac{1}{1+n+n^2}={\tan }^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)$
$={\tan }^{-1}(n+1)-{\tan }^{-1}n$
so, $\sum _{n=1}^{50}\left({\tan }^{-1}(n+1)-{\tan }^{-1}n\right)$
$={\tan }^{-1}51-{\tan }^{-1}1$
$\cot \left(\sum _{n=1}^{50}{\tan }^{-1}\left(\frac{1}{1+n+n^2}\right)\right)=\cot \left({\tan }^{-1}51+{\tan }^{-1}1\right)$
$=\frac{1}{\tan \left({\tan }^{-1}51-{\tan }^{-1}1\right)}=\frac{1+51\times 1}{51-1}$
$=\frac{52}{50}=\frac{26}{25}$