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

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(A) Given the equation: $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$. Let $\cos^{-1} x = A$ and $\cos^{-1} \frac{y}{2} = B$. Then $x = \cos A$ and $

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$4 \sin^2 \alpha$

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(A) Given the equation: $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$. Let $\cos^{-1} x = A$ and $\cos^{-1} \frac{y}{2} = B$. Then $x = \cos A$ and $\frac{y}{2} = \cos B$,so $y = 2 \cos B$. The given equation becomes $A - B = \alpha$. Taking cosine on both sides: $\cos(A - B) = \cos \alpha$. Using the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we get: $x \cdot \frac{y}{2} + \sqrt{1 - x^2} \sqrt{1 - (\frac{y}{2})^2} = \cos \alpha$. $\sqrt{1 - x^2} \sqrt{1 - \frac{y^2}{4}} = \cos \alpha - \frac{xy}{2}$. Squaring both sides: $(1 - x^2)(1 - \frac{y^2}{4}) = (\cos \alpha - \frac{xy}{2})^2$. $1 - \frac{y^2}{4} - x^2 + \frac{x^2 y^2}{4} = \cos^2 \alpha - xy \cos \alpha + \frac{x^2 y^2}{4}$. $1 - x^2 - \frac{y^2}{4} = \cos^2 \alpha - xy \cos \alpha$. Multiply the entire equation by $4$: $4 - 4x^2 - y^2 = 4 \cos^2 \alpha - 4xy \cos \alpha$. Rearranging the terms: $4x^2 - 4xy \cos \alpha + y^2 = 4 - 4 \cos^2 \alpha$. $4x^2 - 4xy \cos \alpha + y^2 = 4(1 - \cos^2 \alpha)$. Since $1 - \cos^2 \alpha = \sin^2 \alpha$,we get: $4x^2 - 4xy \cos \alpha + y^2 = 4 \sin^2 \alpha$.

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

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