Find the value of $4{\tan ^{ – 1}}\frac{1}{5} – {\tan ^{ – 1}}\frac{1}{{239}}$.
[NCERT,Exemplar.2.3,Q.17,Page.37]

We have, $4{\tan ^{ – 1}}\frac{1}{5} – {\tan ^{ – 1}}\frac{1}{{239}}$

$ = 2 \cdot 2{\tan ^{ – 1}}\frac{1}{5} – {\tan ^{ – 1}}\frac{1}{{239}}$

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

$ = 2 \cdot \left[ {{{\tan }^{ – 1}}\left( {\frac{{\frac{2}{5}}}{{1 – \frac{1}{{25}}}}} \right)} \right] – {\tan ^{ – 1}}\frac{1}{{239}}$

$ = 2 \cdot \left[ {{{\tan }^{ – 1}}\left( {\frac{{2/5}}{{24/25}}} \right)} \right] – {\tan ^{ – 1}}\frac{1}{{239}}$

$ = 2{\tan ^{ – 1}}\frac{5}{{12}} – {\tan ^{ – 1}}\frac{1}{{239}}$

$ = {\tan ^{ – 1}}\frac{{2 \cdot \frac{5}{{12}}}}{{1 – {{\left( {\frac{5}{{12}}} \right)}^2}}} – {\tan ^{ – 1}}\frac{1}{{239}}$

$ = {\tan ^{ – 1}}\left( {\frac{{\frac{5}{6}}}{{1 – \frac{{25}}{{144}}}}} \right) – {\tan ^{ – 1}}\frac{1}{{239}}$

$ = {\tan ^{ – 1}}\left( {\frac{{144 \times 5}}{{119 \times 6}}} \right) – {\tan ^{ – 1}}\frac{1}{{239}}$
$ = {\tan ^{ – 1}}\left( {\frac{{120}}{{119}}} \right) – {\tan ^{ – 1}}\frac{1}{{239}}$

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

$ = {\tan ^{ – 1}}\left( {\frac{{120 \times 239 – 119}}{{119 \times 239 + 120}}} \right)$

$ = {\tan ^{ – 1}}\left[ {\frac{{28680 – 119}}{{28441 + 120}}} \right] = {\tan ^{ – 1}}\frac{{28561}}{{28561}}$

$ = {\tan ^{ – 1}}(1) = {\tan ^{ – 1}}\left( {\tan \frac{\pi }{4}} \right) = \frac{\pi }{4}$

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