Prove that  ${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{7}{25}={\sin }^{-1}\frac{117}{125}$.

Let ${\sin }^{-1}\frac{4}{5}=\alpha$ 

${\sin }^{-1}\frac{4}{5}=\alpha \Rightarrow \sin \alpha =\frac{4}{5}$ then $\cos \alpha =\frac{3}{5}$

${\sin }^{-1}\frac{7}{25}=\beta \Rightarrow \sin \beta =\frac{7}{25}$ then $\cos \beta =\frac{24}{25}$

$\sin (\alpha +\beta )=\sin \alpha \cdot \cos \beta +\cos \alpha \cdot \sin \beta$

$=\frac{4}{5}\cdot \frac{24}{25}+\frac{3}{5}\cdot \frac{7}{25}=\frac{96}{125}+\frac{21}{125}=\frac{117}{125}$

$\therefore$$\alpha +\beta ={\sin }^{-1}\left(\frac{117}{125}\right)$

i.e., ${\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{7}{25}\right)={\sin }^{-1}\left(\frac{117}{125}\right)$