$2{\sin ^{ - 1}}\frac{3}{5} = {\tan ^{ - 1}}\frac{{24}}{7}$

Let ${\sin ^{ - 1}}\frac{3}{5} = x$
$\Rightarrow$ $\frac{3}{5} = \sin x \Rightarrow \tan x = \frac{3}{4} \Rightarrow x = {\tan ^{ - 1}}\frac{3}{4}$

$\Rightarrow$ $2{\sin ^{ - 1}}\frac{3}{5} = 2{\tan ^{ - 1}}\frac{3}{4} = {\tan ^{ - 1}}\left( {\frac{{2 \times \frac{3}{4}}}{{1 - {{\left( {\frac{3}{4}} \right)}^2}}}} \right)$

$= {\tan ^{ - 1}}\left( {\frac{{\frac{3}{2}}}{{1 - \frac{9}{{16}}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{\frac{3}{2}}}{{\frac{7}{{16}}}}} \right) = {\tan ^{ - 1}}\left( {\frac{3}{2} \times \frac{{16}}{7}} \right) = {\tan ^{ - 1}}\left( {\frac{{24}}{7}} \right)$