If cos ^{-1} x - cos ^{-1} frac{y}{3} = alpha | vedclass

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(A) Given $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$. Using the identity $\cos ^{-1} a - \cos ^{-1} b = \cos ^{-1} (ab + \sqrt{1-a^2} \sqrt{1-b^

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$9 \sin ^2 \alpha$

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(A) Given $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$. Using the identity $\cos ^{-1} a - \cos ^{-1} b = \cos ^{-1} (ab + \sqrt{1-a^2} \sqrt{1-b^2})$,we have: $\cos ^{-1} \left( \frac{xy}{3} + \sqrt{1-x^2} \sqrt{1-\frac{y^2}{9}} \right) = \alpha$ $\frac{xy}{3} + \frac{\sqrt{1-x^2} \sqrt{9-y^2}}{3} = \cos \alpha$ $xy + \sqrt{1-x^2} \sqrt{9-y^2} = 3 \cos \alpha$ $xy - 3 \cos \alpha = -\sqrt{1-x^2} \sqrt{9-y^2}$ Squaring both sides: $(xy - 3 \cos \alpha)^2 = (1-x^2)(9-y^2)$ $x^2y^2 - 6xy \cos \alpha + 9 \cos ^2 \alpha = 9 - y^2 - 9x^2 + x^2y^2$ $9x^2 - 6xy \cos \alpha + y^2 = 9 - 9 \cos ^2 \alpha$ $9x^2 - 6xy \cos \alpha + y^2 = 9(1 - \cos ^2 \alpha) = 9 \sin ^2 \alpha$.

If $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$,where $-1 \leq x \leq 1$,$-3 \leq y \leq 3$,and $x \leq \frac{y}{3}$,then for all $x, y$,$9x^2 - 6xy \cos \alpha + y^2$ is equal to

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