1. Tardigrade
  2. Question
  3. Mathematics
  4. The value of cot displaystyle ∑n=123 cot-1(1+ displaystyle ∑k=1n 2k ) is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The value of cot $\bigg \{ \displaystyle \sum_{n=1}^{23} \, cot^{-1}\bigg(1+\displaystyle \sum_{k=1}^n 2k \bigg)\bigg \} $ is

JEE MainJEE Main 2013Inverse Trigonometric Functions

Solution:

We have, cot $\bigg [ \displaystyle \sum_{n=1}^{23} \, cot^{-1}\bigg(1+\displaystyle \sum_{k=1}^n 2k \bigg)\bigg ]$
$\Rightarrow \, \, \, $ cot $\bigg [ \displaystyle \sum_{n=1}^{23} \, cot^{-1} (1+2+4+6+8+......+2n)\bigg]$
$\Rightarrow \, \, \, $ cot $\bigg [ \displaystyle \sum_{n=1}^{23} \, cot^{-1} \{1+n(n+1)\}\bigg]$
$\Rightarrow \, \, \, $ cot $\bigg [ \displaystyle \sum_{n=1}^{23} \, tan^{-1} \frac{1}{1+n(n+1)}\bigg]$
$\Rightarrow \, \, \, $ cot $\bigg [ \displaystyle \sum_{n=1}^{23} \, tan^{-1} \bigg\{ \frac{(n+1)-n}{1+n(n+1)}\bigg \}\bigg]$
$\Rightarrow \, \, \, $ cot $\bigg [ \displaystyle \sum_{n=1}^{23} \, ( tan^{-1} (n+1)-tan^{-1}ln \, n) \bigg]$
$\Rightarrow \, \, \, $ cot $ [ ( tan^{-1} 2 -tan^{-1} 1)+(tan^{-1} 3 -tan^{-1} 2) ]$
$ \, \, \, \, \, \, \, \, \, \, +(tan^{-1} 4 -tan^{-1} 3)] + ...+ (tan^{-1} 24 - tan^{-1}23)]$
$\Rightarrow \, \, \, $ cot $(tan^{-1}24 - tan^{-1} 1)$
$\Rightarrow \, \, \, $ cot $\bigg( tan^{-1}\frac{24-1}{1+24.(1)}\bigg)=cot\bigg(tan^{-1} \frac{23}{25}\bigg)$
$ =cot\bigg(cot^{-1} \frac{25}{23}\bigg)=\frac{25}{23}$