If theta is eliminated from the equations x = | vedclass

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(B) Given equations are $\frac{x}{a} = \cos(	heta - \alpha)$ and $\frac{y}{b} = \cos(	heta - \beta)$.We know that $(\alpha - \beta) = (	heta - \bet

Vedclass · Accepted Answer

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

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(B) Given equations are $\frac{x}{a} = \cos(	heta - \alpha)$ and $\frac{y}{b} = \cos(	heta - \beta)$.We know that $(\alpha - \beta) = (	heta - \beta) - (	heta - \alpha)$.Using the cosine difference formula,$\cos(\alpha - \beta) = \cos((	heta - \beta) - (	heta - \alpha)) = \cos(	heta - \beta) \cos(	heta - \alpha) + \sin(	heta - \beta) \sin(	heta - \alpha)$.Substituting the given values,$\cos(\alpha - \beta) = \frac{y}{b} \cdot \frac{x}{a} + \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}}$.Rearranging,$\cos(\alpha - \beta) - \frac{xy}{ab} = \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}}$.Squaring both sides,$\cos^2(\alpha - \beta) + \frac{x^2y^2}{a^2b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = (1 - \frac{x^2}{a^2})(1 - \frac{y^2}{b^2})$.Expanding the right side,$\cos^2(\alpha - \beta) + \frac{x^2y^2}{a^2b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2} + \frac{x^2y^2}{a^2b^2}$.Canceling $\frac{x^2y^2}{a^2b^2}$ from both sides and rearranging,$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos(\alpha - \beta) = 1 - \cos^2(\alpha - \beta)$.Therefore,the expression is equal to $\sin^2(\alpha - \beta)$.

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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