Find the value of : 4tan^-1(1/5) …

We have, {tex}4{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}}{/tex}
{tex}= 2.2{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}}{/tex}
{tex}= 2\left[ {{{\tan }^{ - 1}}\frac{{\frac{2}{5}}}{{1 - {{\left( {\frac{1}{5}} \right)}^2}}}} \right] - {\tan ^{ - 1}}\frac{1}{{239}}{/tex} {tex}\left[ {\because 2{{\tan }^{ - 1}}x = {{\tan }^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)} \right]{/tex}
{tex} = 2.\left[ {{{\tan }^{ - 1}}\left( {\frac{{\frac{2}{5}}}{{1 - \frac{1}{{25}}}}} \right)} \right] - {\tan ^{ - 1}}\frac{1}{{239}}{/tex}
{tex} = 2.\left[ {{{\tan }^{ - 1}}\left( {\frac{{\frac{2}{5}}}{{\frac{{24}}{{25}}}}} \right)} \right] - {\tan ^{ - 1}}\frac{1}{{239}}{/tex}
{tex}= 2{\tan ^{ - 1}}\frac{5}{{12}} - ta{n^{ - 1}}\frac{1}{{239}}{/tex}
{tex}= {\tan ^{ - 1}}\frac{{2.\frac{5}{{12}}}}{{1 - {{\left( {\frac{5}{{12}}} \right)}^2}}} - {\tan ^{ - 1}}\frac{1}{{239}}{/tex} {tex}\left[ {\because 2{{\tan }^{ - 1}}x = {{\tan }^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)} \right]{/tex}
{tex} = {\tan ^{ - 1}}\left( {\frac{{\frac{5}{6}}}{{1 - \frac{{25}}{{144}}}}} \right) - {\tan ^{ - 1}}\frac{1}{{239}}{/tex}
{tex} = {\tan ^{ - 1}}\left( {\frac{{144 \times 5}}{{119 \times 6}}} \right) - {\tan ^{ - 1}}\frac{1}{{239}}{/tex}
{tex} = {\tan ^{ - 1}}\left( {\frac{{120}}{{199}}} \right) - {\tan ^{ - 1}}\frac{1}{{239}}{/tex}
{tex} = {\tan ^{ - 1}}\left( {\frac{{\frac{{120}}{{119}} - \frac{1}{{239}}}}{{1 + \frac{{120}}{{119}} \cdot \frac{1}{{239}}}}} \right){/tex}{tex}\left[ {\because \;{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right)} \right]{/tex}
{tex} = {\tan ^{ - 1}}\left( {\frac{{120 \times 239 - 119}}{{119 \times 239 \times 120}}} \right){/tex}
{tex} = {\tan ^{ - 1}}\left[ {\frac{{28680 - 119}}{{28441 + 120}}} \right] = {\tan ^{ - 1}}\frac{{28561}}{{28561}}{/tex}
{tex}= {\tan ^{ - 1}}(1) = {\tan ^{ - 1}}\left( {\tan \frac{\pi }{4}} \right) = \frac{\pi }{4}{/tex}