If ${\mathrm{Sin}}^{-1}\frac{\mathrm{x}}{2}+{\mathrm{Sin}}^{-1}\frac{\mathrm{y}}{3}=\mathrm{\alpha }$ then $9{\mathrm{x}}^2+12\mathrm{xy} \mathrm{cos\alpha }+4{\mathrm{y}}^2=$

$$\begin{gathered} \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-{\cos }^{-1}\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\frac{\mathrm{y}}{3}=\mathrm{\alpha } \\ \mathrm{\pi }-\mathrm{\alpha }={\cos }^{-1}\frac{\mathrm{x}}{2}+{\cos }^{-1}\frac{\mathrm{y}}{3} \\ ={\cos }^{-1}\left(\frac{\mathrm{x}}{2}·\frac{\mathrm{y}}{3}-\sqrt{1-\frac{{\mathrm{y}}^2}{9}}\right) \\ -\cos \mathrm{\alpha }=\frac{\mathrm{xy}}{6}-\sqrt{1-\frac{{\mathrm{x}}^2}{4}} \sqrt{1-\frac{{\mathrm{y}}^2}{9}} \\ \frac{\mathrm{xy}}{6}+\cos \mathrm{\alpha }=\sqrt{1-\frac{{\mathrm{x}}^2}{4}} \sqrt{\frac{1-{\mathrm{y}}^2}{9}} \\ \mathrm{S}.\mathrm{B}.\mathrm{S} \\ \frac{{\mathrm{x}}^2{\mathrm{y}}^2}{36}+{\cos }^2\mathrm{\alpha }+\frac{\mathrm{xy}}{3}\cos \mathrm{\alpha }=1-\frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{9}+\frac{{\mathrm{x}}^2{\mathrm{y}}^2}{36} \\ 9{\mathrm{x}}^2+4{\mathrm{y}}^2+12\mathrm{xy} \cos \mathrm{\alpha }=36 {\sin }^2\mathrm{\alpha } \end{gathered}$$