If ${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\frac{\sqrt{1-{\mathrm{y}}^2}}{\mathrm{y}}=\frac{\pi }{3}$ and ${\sin }^{-1}y-{\cos }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)=\frac{\pi }{6}$, then $\frac{5{\sin }^{-1}x}{{\sin }^{-1}y}$ is 

${\tan }^{-1}x+{\cos }^{-1}y=\frac{\pi }{3}\text{ and }{\sin }^{-1}y-{\cot }^{-1}x=\frac{\pi }{6}$

$\Rightarrow {\tan }^{-1}x+{\cos }^{-1}y+{\sin }^{-1}y-{\cot }^{-1}x=\frac{\pi }{2}$

Or

$\Rightarrow {\tan }^{-1}x={\cot }^{-1}x\Rightarrow x=1$

$\text{ also, }{\tan }^{-1}x+{\cos }^{-1}y-{\sin }^{-1}y+{\cot }^{-1}x=\frac{\pi }{6}$

$\Rightarrow \pi -2{\sin }^{-1}y=\frac{\pi }{6}\Rightarrow {\sin }^{-1}y=\frac{5\pi }{12}$

$\therefore \frac{5{\sin }^{-1}x}{{\sin }^{-1}y}=6$