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The value of $\cot\sum\limits_{n=1}^{23}\cot^{-1}\big(1+\sum\limits_{k=1}^n 2k\big)$ is

$(a)\;\large\frac{23}{25}$$\qquad(b)\;\large\frac{25}{23}$$\qquad(c)\;\large\frac{23}{24}\qquad$$(d)\;\large\frac{24}{23}$

1 Answer

$\cot^{-1}\big(1+\sum\limits_{k=1}^n2k\big)=\cot^{-1}[1+n(n+1)]$
$\qquad\qquad\qquad\quad=\tan^{-1}\big[\large\frac{(n+1)-n}{1+(n+1)n}\big]$
$\qquad\qquad\qquad\quad=\tan^{-1}(n+1)-\tan^{-1}n$
$\therefore \sum\limits_{n=1}^{23}[\tan^{-1}(n+1)-\tan^{-1}n]=\tan^{-1}24-\tan^{-1}1$
$\qquad\qquad\qquad\qquad\qquad\quad\;\;=\tan^{-1}\large\frac{23}{25}$
$\therefore \cot\big[\sum\limits_{n=1}^{23}\cot^{-1}\big(1+\sum\limits_{k=1}^n 2k\big)\big]=\cot^{-1}\big[\tan^{-1}\large\frac{23}{25}\big]$
$\Rightarrow \large\frac{25}{23}$
Hence (b) is the correct answer.
answered Oct 16, 2013 by sreemathi.v
 

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