If \(\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1} \alpha\), then \(\alpha=\)

We have \(\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1} \alpha\)
\(
\begin{aligned}
& \therefore \tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right)=\tan ^{-1}\left(\frac{\alpha}{\sqrt{1-\alpha^2}}\right) \\
& \therefore \tan ^{-1}\left[\frac{\left(\frac{3}{4}\right)+\left(\frac{5}{12}\right)}{1-\left(\frac{3}{4}\right)\left(\frac{5}{12}\right)}\right]=\tan ^{-1}\left(\frac{\alpha}{\sqrt{1-\alpha^2}}\right) \\
& \therefore \tan ^{-1}\left[\frac{\left(\frac{14}{12}\right)}{\left(\frac{11}{16}\right)}\right]=\tan ^{-1}\left(\frac{\alpha}{\sqrt{1-\alpha^2}}\right) \\
& \therefore \frac{56}{33}=\frac{\alpha}{\sqrt{1-\alpha^2}} \Rightarrow(56)^2(1-\alpha)^2=(33)^2 \alpha^2 \\
& \alpha^2=\frac{(56)^2}{(56)^2+(33)^2}=\frac{(56)^2}{(65)^2} \Rightarrow \alpha=\frac{56}{65}
\end{aligned}
\)