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

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(C) Given the equation $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$. Using the identity $\cos ^{-1} u - \cos ^{-1} v = \cos ^{-1} (uv + \sqrt{1-u^

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$9\sin ^2 \alpha$

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(C) Given the equation $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$. Using the identity $\cos ^{-1} u - \cos ^{-1} v = \cos ^{-1} (uv + \sqrt{1-u^2}\sqrt{1-v^2})$,we have $\cos ^{-1} (\frac{xy}{3} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}}) = \alpha$. Taking cosine on both sides: $\frac{xy}{3} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}} = \cos \alpha$. Rearranging: $\sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}} = \cos \alpha - \frac{xy}{3}$. Squaring both sides: $(1-x^2)(1-\frac{y^2}{9}) = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha + \frac{x^2 y^2}{9}$. $1 - \frac{y^2}{9} - x^2 + \frac{x^2 y^2}{9} = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha + \frac{x^2 y^2}{9}$. $1 - x^2 - \frac{y^2}{9} = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha$. $1 - \cos^2 \alpha = x^2 - \frac{2xy}{3} \cos \alpha + \frac{y^2}{9}$. $\sin^2 \alpha = x^2 - \frac{2xy}{3} \cos \alpha + \frac{y^2}{9}$. Multiplying by $9$: $9 \sin^2 \alpha = 9x^2 - 6xy \cos \alpha + y^2$.

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 $x, y$,$9 x^2-6 x y \cos \alpha+y^2$ is equal to

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