Inverse Trigonometric Functions — Class 12 Maths Solution

$2{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = {\tan ^{ - 1}}\frac{{\left( {2 \times \frac{1}{2}} \right)}}{{\left\{ {1 - {{\left( {\frac{1}{2}} \right)}^2}} \right\}}} + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)$

$= {\tan ^{ - 1}}\left( {\frac{1}{{3/4}}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)$

$= {\tan ^{ - 1}}\left[ {\frac{{\frac{4}{3} + \frac{1}{7}}}{{1 - \frac{4}{3} \times \frac{1}{7}}}} \right] = {\tan ^{ - 1}}\left( {\frac{{\frac{{31}}{{21}}}}{{\frac{{17}}{{21}}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{31}}{{17}}} \right)$

$\therefore$ $2{\tan ^{ - 1}}\frac{1}{2} + {\tan ^{ - 1}}\frac{1}{7} = {\tan ^{ - 1}}\frac{{31}}{{17}}$