Prove that \( 4\tan^{-1}\left(\frac{1}{5}\right) – \tan^{-1}\left(\frac{1}{239}\right) = \frac{\pi}{4} \)

Solution:

Let

\[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \Rightarrow \tan \theta = \frac{1}{5} \]

Step 1: Find \( \tan(2\theta) \)

\[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} = \frac{2/5}{1 – 1/25} = \frac{2/5}{24/25} = \frac{5}{12} \]

\[ \Rightarrow 2\theta = \tan^{-1}\left(\frac{5}{12}\right) \]

Step 2: Find \( \tan(4\theta) \)

\[ \tan(4\theta) = \frac{2\tan(2\theta)}{1 – \tan^2(2\theta)} = \frac{2 \cdot \frac{5}{12}}{1 – \left(\frac{5}{12}\right)^2} = \frac{10/12}{1 – 25/144} = \frac{10/12}{119/144} \]

\[ = \frac{10}{12} \times \frac{144}{119} = \frac{120}{119} \]

\[ \Rightarrow 4\theta = \tan^{-1}\left(\frac{120}{119}\right) \]

Step 3: Evaluate the expression

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

Using identity:

\[ \tan^{-1}a – \tan^{-1}b = \tan^{-1}\left(\frac{a – b}{1 + ab}\right) \]

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

\[ = \tan^{-1}\left(\frac{\frac{28680 – 119}{28441}}{1 + \frac{120}{28441}}\right) \]

\[ = \tan^{-1}\left(\frac{28561/28441}{28561/28441}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]

Hence,

\[ 4\tan^{-1}\left(\frac{1}{5}\right) – \tan^{-1}\left(\frac{1}{239}\right) = \frac{\pi}{4} \]

Final Answer:

\[ 4\tan^{-1}\left(\frac{1}{5}\right) – \tan^{-1}\left(\frac{1}{239}\right) = \frac{\pi}{4} \]