Question

Find the sum of $cose{c}^{-1}\sqrt{5}+cose{c}^{-1}\sqrt{65}+cose{c}^{-1}\sqrt{325}+....$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{4}$

${\csc }^{-1}\sqrt{5}+{\csc }^{-1}\sqrt{65}+{\csc }^{-1}\sqrt{325}+...$
$={\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{8}+{\tan }^{-1}\frac{1}{18}+...+{\tan }^{-1}\frac{1}{{2n}^2}+...$
$=\underset{n\to \infty }{lim}\sum _{r=1}^n{\tan }^{-1}\frac{1}{{2n}^2}=\underset{n\to \infty }{lim}\sum _{r=1}^n{\tan }^{-1}\left(\frac{(2n+1)-(2n-1)}{1+(2n-1)(2n+1)}\right)={lim}_{n\to \infty }{\sum }_{r=1}^n\left[{\tan }^{-1}(2n+1)-{\tan }^{-1}(2n-1)\right]={lim}_{n\to \infty }[{\tan }^{-1}(2n+1)-{\tan }^{-1}1]$
$={\tan }^{-1}\infty -\frac{\pi }{4}=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$