If cos ^{-1} x - cos ^{-1} frac{y}{2} = alpha | vedclass

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(B) Given $\cos ^{-1} x - \cos ^{-1} \frac{y}{2} = \alpha$. Using the formula $\cos ^{-1} A - \cos ^{-1} B = \cos ^{-1} (AB + \sqrt{1-A^2} \sqrt{1-B^2

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$4 \sin ^2 \alpha$

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(B) Given $\cos ^{-1} x - \cos ^{-1} \frac{y}{2} = \alpha$. Using the formula $\cos ^{-1} A - \cos ^{-1} B = \cos ^{-1} (AB + \sqrt{1-A^2} \sqrt{1-B^2})$,we get: $\cos ^{-1} \left( x \cdot \frac{y}{2} + \sqrt{1-x^2} \sqrt{1-\frac{y^2}{4}} \right) = \alpha$. Taking cosine on both sides: $\frac{xy}{2} + \frac{\sqrt{(1-x^2)(4-y^2)}}{2} = \cos \alpha$. $\sqrt{(1-x^2)(4-y^2)} = 2 \cos \alpha - xy$. Squaring both sides: $(1-x^2)(4-y^2) = (2 \cos \alpha - xy)^2$. $4 - y^2 - 4x^2 + x^2y^2 = 4 \cos ^2 \alpha - 4xy \cos \alpha + x^2y^2$. Subtracting $x^2y^2$ from both sides: $4 - y^2 - 4x^2 = 4 \cos ^2 \alpha - 4xy \cos \alpha$. Rearranging the terms: $4x^2 - 4xy \cos \alpha + y^2 = 4 - 4 \cos ^2 \alpha$. Since $1 - \cos ^2 \alpha = \sin ^2 \alpha$,we have: $4x^2 - 4xy \cos \alpha + y^2 = 4 \sin ^2 \alpha$.

If $\cos ^{-1} x - \cos ^{-1} \frac{y}{2} = \alpha$,where $-1 \leq x \leq 1, -2 \leq y \leq 2, x \leq \frac{y}{2}$,then for all $x, y$,$4x^2 - 4xy \cos \alpha + y^2$ is equal to

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