The angle which is equivalent to $4{\tan }^{-1}\left(\frac{1}{5}\right)-{\tan }^{-1}\left(\frac{1}{239}\right)$  is equal to 

$\alpha =2{\tan }^{-1}\left(\frac{2/5}{1-1/25}\right)-{\tan }^{-1}\left(\frac{1}{239}\right)$               $[2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right),-1<x<1]$                 
 $=2{\tan }^{-1}\left(\frac{5}{12}\right)-{\tan }^{-1}\left(\frac{1}{239}\right)={\tan }^{-1}\left(\frac{5/6}{1-25/144}\right)-{\tan }^{-1}\left(\frac{1}{239}\right)$
 $\alpha ={\tan }^{-1}\left(\frac{120}{119}\right)-{\tan }^{-1}\left(\frac{1}{239}\right)$               ……(1)
But  $\frac{\pi }{4}+{\tan }^{-1}\left(\frac{1}{239}\right)={\tan }^{-1}1+{\tan }^{-1}\left(\frac{1}{239}\right)={\tan }^{-1}\left(\frac{240/239}{1-1/239}\right)={\tan }^{-1}\left(\frac{120}{119}\right)$
 $\therefore {\tan }^{-1}\left(\frac{120}{119}\right)-{\tan }^{-1}\left(\frac{1}{239}\right)=\frac{\pi }{4}$