Question

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