Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

We have, $\sum _{m=1}^n{\tan }^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$

$=\sum _{m=1}^n{\tan }^{-1}\left(\frac{\left(m^2+m+1\right)-\left(m^2-m+1\right)}{1+\left(m^2+m+1\right)\left(m^2-m+1\right)}\right)$

$=\sum _{m=1}^n\left[{\tan }^{-1}\left(m^2+m+1\right)-{\tan }^{-1}\left(m^2-m+1\right)\right]$

$=\left({\tan }^{-1}3-{\tan }^{-1}1\right)+\left({\tan }^{-1}7-{\tan }^{-1}3\right)+\left({\tan }^{-1}\left(13\right)-{\tan }^{-1}7\right)+\dots +\left({\tan }^{-1}\left(n^2+n+1\right)\right.\left.-{\tan }^{-1}\left(n^2-n+1\right)\right)$

$={\tan }^{-1}\left(n^2+n+1\right)-{\tan }^{-1}1$

$={\tan }^{-1}\left(\frac{n^2+n+1-1}{1+\left(n^2+n+1\right)}\right)={\tan }^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$