Question

${\cot }^{-1}\left(2.1^2\right)+{\cot }^{-1}\left(2.2^2\right)+{\cot }^{-1}\left(2.3^2\right)+\dots$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{5}$

Answer 1

Answer: (A)

${\cot }^{-1}\left(2.1^2\right)+{\cot }^{-1}\left(2.2^2\right)+{\cot }^{-1}\left(2.3^2\right)+\dots$
$=\sum _{r=1}^{\infty }{\cot }^{-1}\left(2\cdot r^2\right)=\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{2r^2}\right)$
$=\sum _{r=1}^{\infty }{\tan }^{-1}\left[\frac{(1+2r)+(1-2r)}{1-(1+2r)(1-2r)}\right]$
$=\sum _{r=1}^{\infty }\left[{\tan }^{-1}(1+2r)+{\tan }^{-1}(1-2r)\right]$
$={\tan }^{-1}3-{\tan }^{-1}1+{\tan }^{-1}5-{\tan }^{-1}3+{\tan }^{-1}7$
$-{\tan }^{-1}5+\dots .+{\tan }^{-1}\infty$
$=-\frac{\pi }{4}+\frac{\pi }{2}=\frac{\pi }{4}$