The value of

$\cot \left(\sum_{n=1}^{19} \cot ^{-1}(1+n(n+1))\right.$

$\cot \left(\sum_{n=1}^{19} \cot ^{-1}\left(n^{2}+n+1\right)\right)=\cot \left(\sum_{n=1}^{19} \tan ^{-1} \frac{1}{1+n(n+1)}\right)$

$\sum_{n=1}^{19}\left(\tan ^{-1}(n+1)-\tan ^{-1} n\right)$

$\cot \left(\tan ^{-1} 20-\tan ^{-1} 1\right)=\frac{\cot A \cot \beta+1}{\cot \beta-\cot A}$

(Where $\tan \mathrm{A}=20, \tan \mathrm{B}=1$ )

$\frac{1\left(\frac{1}{20}\right)+1}{1-\frac{1}{20}}=\frac{21}{19}$

$\therefore$ Option (3)