If $cos^{-1}\frac{x}{2}+cos^{-1}\frac{y}{3}=\theta $, then $9x^2 - 12 xy cos \theta + 4y^2 $ is equal to
Answer & explanation
Correct answer: option 3
We have,
$cos^{-1}\frac{x}{2}+cos^{-1}\frac{y}{3}=\theta $
$⇒ cos^{-1}\begin{Bmatrix}\frac{xy}{6}-\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{9}}\end{Bmatrix}=\theta $
$⇒ xy - \sqrt{4-x^2}\sqrt{9-y^2}= 6 cos \theta $
$⇒ - 12 xy cos \theta + 36 cos^2 \theta - 36 - 4y^2 - 9x^2 $
$⇒ 9x^2 + 4y^2 - 12 xy cos \theta = 36 sin^2 \theta $