Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$\sum _{r=1}^{50}{\tan }^{-1}\left(\frac{1}{2r^2}\right)=\sum _{r=1}^{50}{\tan }^{-1}\left(\frac{2}{4r^2}\right)$

$=\sum _{r=1}^{50}{\tan }^{-1}\left(\frac{2}{1+4r^2-1}\right)$

$=\sum _{r=1}^{50}{\tan }^{-1}\left(\frac{2}{1+\left(2r-1\right)\left(2r+1\right)}\right)$

$=\sum _{r=1}^{50}{\tan }^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right)$

$=\sum _{r=1}^{50}[{\tan }^{-1}\left(2r+1\right)-{\tan }^{-1}\left(2r-1\right)]$

So, $\sum _{r=1}^{50}{\tan }^{-1}\left(\frac{1}{2r^2}\right)={\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(5\right)-{\tan }^{-1}\left(3\right)+{\tan }^{-1}\left(7\right)-{\tan }^{-1}\left(5\right)...........{\tan }^{-1}\left(101\right)-{\tan }^{-1}\left(99\right)={\tan }^{-1}\left(101\right)-{\tan }^{-1}\left(1\right)$

$={\tan }^{-1}\left(\frac{101-1}{1+101}\right)$

$={\tan }^{-1}\left(\frac{100}{102}\right)=p$

So, $\tan p=\frac{100}{102}=\frac{50}{51}$