Question

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

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

Answer 1

The correct option is C $\frac{21}{19}$

Given
$\cot \left(\sum _{n=1}^{19}{\cot }^{-1}(1+\sum _{p=1}^{n}2p)\right)$
$=\cot \left(\sum _{n=1}^{19}{\cot }^{-1}\right(1+n(n+1)\left)\right)$
$=\cot \left(\sum _{n=1}^{19}{\tan }^{-1}\frac{1}{1+n(n+1)}\right)$
$=\cot \left(\sum _{n=1}^{19}{\tan }^{-1}\frac{(n+1)-n}{1+n(n+1)}\right)$
$=\cot \left(\sum _{n=1}^{19}{\tan }^{-1}\right(n+1)-{\tan }^{-1}n)$
$=\cot ({\tan }^{-1}20-{\tan }^{-1}1)$
$=\cot \left({\tan }^{-1}\frac{20-1}{1+20\cdot 1}\right)$
$=\cot \left({\tan }^{-1}\frac{19}{21}\right)$
$=\cot \left({\cot }^{-1}\frac{21}{19}\right)$

$=\frac{21}{19}$