If cos^{-1} left( frac{x}{2} right) + cos^{-1 | vedclass

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(A) Given: $\cos^{-1} \left( \frac{x}{2} ight) + \cos^{-1} \left( \frac{y}{3} ight) = 	heta$ Using the identity $\cos^{-1} A + \cos^{-1} B = \cos

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$36 \sin^{2} 	heta$

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(A) Given: $\cos^{-1} \left( \frac{x}{2} ight) + \cos^{-1} \left( \frac{y}{3} ight) = 	heta$ Using the identity $\cos^{-1} A + \cos^{-1} B = \cos^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} ight)$,we get: $\cos^{-1} \left\{ \left( \frac{x}{2} ight) \left( \frac{y}{3} ight) - \sqrt{1 - \frac{x^{2}}{4}} \sqrt{1 - \frac{y^{2}}{9}} ight\} = 	heta$ $\Rightarrow \frac{xy}{6} - \sqrt{\frac{4-x^2}{4}} \sqrt{\frac{9-y^2}{9}} = \cos 	heta$ $\Rightarrow \frac{xy}{6} - \frac{\sqrt{4-x^2} \sqrt{9-y^2}}{6} = \cos 	heta$ $\Rightarrow xy - \sqrt{4-x^2} \sqrt{9-y^2} = 6 \cos 	heta$ $\Rightarrow xy - 6 \cos 	heta = \sqrt{4-x^2} \sqrt{9-y^2}$ Squaring both sides: $(xy - 6 \cos 	heta)^2 = (4-x^2)(9-y^2)$ $x^2 y^2 - 12xy \cos 	heta + 36 \cos^2 	heta = 36 - 4y^2 - 9x^2 + x^2 y^2$ Canceling $x^2 y^2$ from both sides: $-12xy \cos 	heta + 36 \cos^2 	heta = 36 - 9x^2 - 4y^2$ Rearranging the terms: $9x^2 + 4y^2 - 12xy \cos 	heta = 36 - 36 \cos^2 	heta$ $9x^2 + 4y^2 - 12xy \cos 	heta = 36(1 - \cos^2 	heta)$ $9x^2 + 4y^2 - 12xy \cos 	heta = 36 \sin^2 	heta$

If $\cos^{-1} \left( \frac{x}{2} \right) + \cos^{-1} \left( \frac{y}{3} \right) = \theta$,then $9 x^{2} - 12 x y \cos \theta + 4 y^{2} =$

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