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(B) Given equations are $x = a \cos(	heta - \alpha)$ and $y = b \cos(	heta - \beta)$.This implies $\frac{x}{a} = \cos(	heta - \alpha)$ and $\frac{y

Vedclass · Accepted Answer

$\sin^2(\alpha - \beta)$

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(B) Given equations are $x = a \cos(	heta - \alpha)$ and $y = b \cos(	heta - \beta)$.This implies $\frac{x}{a} = \cos(	heta - \alpha)$ and $\frac{y}{b} = \cos(	heta - \beta)$.Consider the identity $\cos(\alpha - \beta) = \cos((	heta - \beta) - (	heta - \alpha))$.Using the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we get:$\cos(\alpha - \beta) = \cos(	heta - \beta) \cos(	heta - \alpha) + \sin(	heta - \beta) \sin(	heta - \alpha)$.Substituting the given values:$\cos(\alpha - \beta) = \frac{y}{b} \cdot \frac{x}{a} + \sqrt{1 - \frac{y^2}{b^2}} \cdot \sqrt{1 - \frac{x^2}{a^2}}$.Rearranging the terms:$\cos(\alpha - \beta) - \frac{xy}{ab} = \sqrt{1 - \frac{y^2}{b^2}} \cdot \sqrt{1 - \frac{x^2}{a^2}}$.Squaring both sides:$\left(\cos(\alpha - \beta) - \frac{xy}{ab}ight)^2 = \left(1 - \frac{y^2}{b^2}ight) \left(1 - \frac{x^2}{a^2}ight)$.$\cos^2(\alpha - \beta) + \frac{x^2y^2}{a^2b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2y^2}{a^2b^2}$.Canceling $\frac{x^2y^2}{a^2b^2}$ from both sides and rearranging:$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = 1 - \cos^2(\alpha - \beta)$.Since $1 - \cos^2(	heta) = \sin^2(	heta)$,the expression equals $\sin^2(\alpha - \beta)$.

If $\theta$ is eliminated from the equations $x = a \cos(\theta - \alpha)$ and $y = b \cos(\theta - \beta)$,then $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos(\alpha - \beta)$ is equal to

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