The distance between the points (a cos alpha, | vedclass

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(D) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Substituting the given points

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$2a \sin \frac{\alpha - \beta}{2}$

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(D) The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Substituting the given points:$d = \sqrt{(a \cos \beta - a \cos \alpha)^2 + (a \sin \beta - a \sin \alpha)^2}$$d = \sqrt{a^2(\cos^2 \beta + \cos^2 \alpha - 2 \cos \alpha \cos \beta) + a^2(\sin^2 \beta + \sin^2 \alpha - 2 \sin \alpha \sin \beta)}$$d = a \sqrt{(\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta)}$Using trigonometric identities $\cos^2 	heta + \sin^2 	heta = 1$ and $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$:$d = a \sqrt{1 + 1 - 2 \cos(\alpha - \beta)}$$d = a \sqrt{2(1 - \cos(\alpha - \beta))}$Using $1 - \cos 	heta = 2 \sin^2(	heta/2)$:$d = a \sqrt{2 	imes 2 \sin^2 \left(\frac{\alpha - \beta}{2}ight)}$$d = 2a \left| \sin \left(\frac{\alpha - \beta}{2}ight) ight|$

The distance between the points $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$ is

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