Question

The value of ${\sum }_{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{r^2+5r+7}\right)$ is equal to

• A. ${\tan }^{-1}3$
• B. $\frac{\pi }{4}$
• C. ${\sin }^{-1}\frac{1}{\sqrt{10}}$
• D. ${\cot }^{-1}2$

Answer 1

Answer: (C)

${\tan }^{-1}\left(\frac{1}{r^2+5r+7}\right)={\tan }^{-1}\left[\frac{(r+3)-(r+2)}{1+(r+3)(r+2)}\right]$
$={\tan }^{-1}(r+3)-{\tan }^{-1}(r+2)$
$\sum _{r=1}^{\infty }{\tan }^{-1}\left(\frac{1}{r^2+5r+7}\right)={\tan }^{-1}4-{\tan }^{-1}3+{\tan }^{-1}5-{\tan }^{-1}4+\dots \dots \dots +{\tan }^{-1}(\infty )$
$=\frac{\pi }{2}-{\tan }^{-1}3={\cot }^{-1}3={\sin }^{-1}\left(\frac{1}{\sqrt{10}}\right)$