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(A) Let the points be $A = (a \cos 	heta, a \sin 	heta)$ and $B = (a \cos \phi, a \sin \phi)$.Given the distance $AB = 2a$.Using the distance formul

Vedclass · Accepted Answer

$2n\pi \pm \pi + \phi, n \in Z$

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(A) Let the points be $A = (a \cos 	heta, a \sin 	heta)$ and $B = (a \cos \phi, a \sin \phi)$.Given the distance $AB = 2a$.Using the distance formula:$AB^2 = (a \cos 	heta - a \cos \phi)^2 + (a \sin 	heta - a \sin \phi)^2 = (2a)^2$$a^2(\cos^2 	heta + \cos^2 \phi - 2 \cos 	heta \cos \phi + \sin^2 	heta + \sin^2 \phi - 2 \sin 	heta \sin \phi) = 4a^2$$a^2(1 + 1 - 2(\cos 	heta \cos \phi + \sin 	heta \sin \phi)) = 4a^2$$2a^2(1 - \cos(	heta - \phi)) = 4a^2$$1 - \cos(	heta - \phi) = 2$$\cos(	heta - \phi) = -1$Since $\cos(	heta - \phi) = -1$,we have $	heta - \phi = (2n + 1)\pi = 2n\pi + \pi$ for $n \in Z$.Therefore,$	heta = 2n\pi + \pi + \phi$.

If the distance between the points $(a \cos \theta, a \sin \theta)$ and $(a \cos \phi, a \sin \phi)$ is $2a$,then $\theta$ is equal to

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