MCQ
$\cot \left[ \sum\limits_{n=1}^{23}{{{\cot }^{-1}}\left[ 1+\sum\limits_{k=1}^{n}{2k} \right]} \right]=......$
- A$\frac{23}{25}$
- ✓$\frac{25}{23}$
- C$\frac{23}{24}$
- D$\frac{24}{23}$
$cot \left[\sum_{n = 1}^{23}cot^{-1}\left[1 + \sum_{k = 1}^n 2k \right] \right]$
$= cot \left[\sum_{n = 1}^{23} cot^{-1} (1 + n^2 + n)\right] \ \ \ \ \ \ \ \left(\therefore \ \sum n = \frac{n}{2} (n + 1)\right)$
$= cot \left[\sum_{n = 1}^{23}tan^{-1} \frac{1}{1 + n (n + 1)}\right]$
$= cot \left[\sum_{n = 1}^{23}tan^{-1} \left(\frac{(1 + n) - n}{1 + n (1 + n)}\right)\right]$
$= cot \left[\sum_{n = 1}^{23} \left(tan^{-1} (1 + n) - tan^{-1} n\right)\right]$
$= cot \left[\left(tan^{-1}2 - tan^{-1}1\right) + \left(tan^{-1}3 - tan^{-1}2\right)+ ...... + \left(tan^{-1}24 - tan^{-1} 23\right)\right]$
$= cot \left[tan^{-1}24 - tan^{-1}1\right]$
$= cot \left[tan^{-1} \left(\frac{24 - 1}{1 + 24}\right)\right]$
$= cot \left(tan^{-1} \frac{23}{25}\right)$
$= \frac{25}{23}$
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