Question

The value of $\cot \left(\sum _{n=1}^{19}{\cot }^{-1}\left(1+\sum _{p=1}^{n}2p\right)\right)$ is :

• A. $\frac{22}{23}$
• B. $\frac{23}{22}$
• C. $\frac{21}{19}$
• D. $\frac{19}{21}$

Answer 1

Answer: (C)

$\cot \left({\sum }_{n=1}^{19}{\cot }^{-1}\left(1+n\left(n+1\right)\right)\right)$
$\cot \left({\sum }_{n=1}^{19}{\cot }^{-1}\left(n^2+n+1\right)\right)=\cot \left({\sum }_{n=1}^{19}{\tan }^{-1}\frac{1}{1+n\left(n+1\right)}\right)$
${\sum }_{n=1}^{19}\left({\tan }^{-1}\left(n+1\right)-{\tan }^{-1}n\right)$
$\cot \left({\tan }^{-1}20-{\tan }^{-1}1\right)=\frac{\cot A\cot \beta +1}{\cot \beta -\cot A}$
(Where $\tan A=20,\tan B=1$ )
$\frac{1\left(\frac{1}{20}\right)+1}{1-\frac{1}{20}}=\frac{21}{19}$