Solution

We have: \[ 4\tan^{-1}\left(\frac{1}{5}\right) – \tan^{-1}\left(\frac{1}{239}\right) \]

\[ = 2 \times 2\tan^{-1}\left(\frac{1}{5}\right) – \tan^{-1}\left(\frac{1}{239}\right) \]

Using the identity: \[ 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]

\[ 2\tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{2/5}{1-1/25}\right) \]

\[ = \tan^{-1}\left(\frac{2/5}{24/25}\right) = \tan^{-1}\left(\frac{5}{12}\right) \]

Hence, \[ 4\tan^{-1}\left(\frac{1}{5}\right) = 2\tan^{-1}\left(\frac{5}{12}\right) \]

Again using the identity: \[ 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]

\[ 2\tan^{-1}\left(\frac{5}{12}\right) = \tan^{-1}\left(\frac{5/6}{1-25/144}\right) \]

\[ = \tan^{-1}\left(\frac{120}{119}\right) \]

Now, \[ \tan^{-1}\left(\frac{120}{119}\right) – \tan^{-1}\left(\frac{1}{239}\right) \]

\[ = \tan^{-1}\left( \frac{\frac{120}{119}-\frac{1}{239}} {1+\frac{120}{119}\cdot\frac{1}{239}} \right) \]

\[ = \tan^{-1}\left( \frac{120\times239-119}{119\times239+120} \right) \]

\[ = \tan^{-1}\left(\frac{28561}{28561}\right) = \tan^{-1}(1) \]

\[ \boxed{\frac{\pi}{4}} \]