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

Let $\sin ^{-1} \frac{3}{5} = x$. Then $\sin x = \frac{3}{5}$.
Since $\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Therefore,$\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{4/5} = \frac{3}{4}$.
This implies $x = \tan ^{-1} \frac{3}{4}$,so $\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{3}{4}$.
Now,consider the $L$.$H$.$S$.:
$2 \sin ^{-1} \frac{3}{5} = 2 \tan ^{-1} \frac{3}{4}$.
Using the formula $2 \tan ^{-1} A = \tan ^{-1} \left( \frac{2A}{1 - A^2} \right)$:
$2 \tan ^{-1} \frac{3}{4} = \tan ^{-1} \left( \frac{2 \times \frac{3}{4}}{1 - (\frac{3}{4})^2} \right)$.
$= \tan ^{-1} \left( \frac{3/2}{1 - 9/16} \right) = \tan ^{-1} \left( \frac{3/2}{7/16} \right)$.
$= \tan ^{-1} \left( \frac{3}{2} \times \frac{16}{7} \right) = \tan ^{-1} \frac{24}{7}$.
Thus,$L$.$H$.$S$. = $R$.$H$.$S$.