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

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

Vedclass · Accepted Answer

$36 \sin ^2 	heta$

Vedclass · Answer

(A) Given,$\cos ^{-1}\left(\frac{x}{2}ight)+\cos ^{-1}\left(\frac{y}{3}ight)=	heta$. Using the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} ight)$,we have: $\cos ^{-1} \left( \frac{xy}{6} - \sqrt{1-\frac{x^2}{4}} \sqrt{1-\frac{y^2}{9}} ight) = 	heta$. Taking $\cos$ on both sides: $\frac{xy}{6} - \cos 	heta = \sqrt{\left(1-\frac{x^2}{4}ight)\left(1-\frac{y^2}{9}ight)}$. Squaring both sides: $\left( \frac{xy}{6} - \cos 	heta ight)^2 = \left( 1 - \frac{x^2}{4} ight) \left( 1 - \frac{y^2}{9} ight)$. $\frac{x^2y^2}{36} - \frac{xy \cos 	heta}{3} + \cos^2 	heta = 1 - \frac{y^2}{9} - \frac{x^2}{4} + \frac{x^2y^2}{36}$. Multiplying by $36$: $x^2y^2 - 12xy \cos 	heta + 36 \cos^2 	heta = 36 - 4y^2 - 9x^2 + x^2y^2$. Rearranging the terms: $9x^2 - 12xy \cos 	heta + 4y^2 = 36 - 36 \cos^2 	heta$. Since $1 - \cos^2 	heta = \sin^2 	heta$,we get: $9x^2 - 12xy \cos 	heta + 4y^2 = 36 \sin^2 	heta$.

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

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