Prove

Let $\sin ^{-1} \frac{3}{5}=x$. Then, $\sin x=\frac{3}{5}$.

$\Rightarrow \cos x=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5}$

$\therefore \tan x=\frac{3}{4}$

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

Now, we have:

L.H.S. $=2 \sin ^{-1} \frac{3}{5}=2 \tan ^{-1} \frac{3}{4}$

$=\tan ^{-1}\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)$      $\left[2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]$ 

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

$=\tan ^{-1} \frac{24}{7}=$ R.H.S.