If cos - 1 x 2 + cos - 1 y 3 = θ , then 9 x 2 - 12 x y cos θ + 4 y 2 =

Given:

${\cos }^{-1}\left(\frac{x}{2}\right)+{\cos }^{-1}\left(\frac{y}{3}\right)=\theta$

$\Rightarrow {\cos }^{-1}\left\{\left(\frac{x}{2}\right)\left(\frac{y}{3}\right)-\left(\sqrt{1-\frac{x^2}{4}}\right)\left(\sqrt{1-\frac{y^2}{9}}\right)\right\}=\theta$

$\Rightarrow \left\{\left(\frac{x}{2}\right)\left(\frac{y}{3}\right)-\left(\sqrt{\frac{4-x^2}{4}}\right)\left(\sqrt{\frac{9-y^2}{9}}\right)\right\}=\cos \theta$

$\Rightarrow \left\{\frac{xy}{6}-\left(\frac{\sqrt{4-x^2}}{2}\right)\left(\frac{\sqrt{9-y^2}}{3}\right)\right\}=\cos \theta$

$\Rightarrow xy-\left(\sqrt{4-x^2}\right)·\left(\sqrt{9-y^2}\right)=6\cos \theta$

$\Rightarrow \left(\sqrt{4-x^2}\right)·\left(\sqrt{9-y^2}\right)=xy-6\cos \theta$

Squaring both sides, we get

$\left(4-x^2\right)·\left(9-y^2\right)=36{\cos }^2\theta +x^2{y}^2-12xy\cos \theta$

$\Rightarrow 36-9x^2-4y^2+x^2{y}^2=36{\cos }^2\theta +x^2{y}^2-12xy\cos \theta$

$\Rightarrow 36-36{\cos }^2\theta =9x^2+4y^2-12xy\cos \theta$

$\Rightarrow 9x^2+4y^2-12xy\cos \theta =36{\sin }^2\theta$