If \(cos^{-1} \frac{x}{2}+cos^{-1} \frac{y}{3}=\phi\), then \(9x^2-12xy cos\phi+4y^2\) is
\[ \begin{aligned} & \text{Given that } \cos^{-1} \frac{x}{2} + \cos^{-1} \frac{y}{3} = \phi, \\ & \Rightarrow \cos^{-1} \left( \frac{xy}{6} - \sqrt{1 - \frac{x^2}{4}} \sqrt{1 - \frac{y^2}{9}} \right) = \phi, \\ & \Rightarrow \frac{xy}{6} - \frac{\sqrt{(4 - x^2)(9 - y^2)}}{6} = \cos \phi, \\ & \Rightarrow (4 - x^2)(9 - y^2) = (6 \cos \phi - xy)^2, \\ & \Rightarrow 9x^2 - 12xy \cos \phi + 4y^2 = 36 \sin^2 \phi. \end{aligned} \]
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