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The polar equation of the line perpendicular to the line $\sin \theta- \cos \theta=\large\frac{1}{r}$ and passing through the point $\bigg(2, \large\frac{\pi}{6}\bigg)$ is

$(a)\;\sin \theta + \cos \theta=\frac{\sqrt 3+1}{r} \qquad (b)\;\sin \theta - \cos \theta=\frac{\sqrt 3+1}{r} \qquad (c)\;\sin \theta + \cos \theta=\frac{\sqrt 3-1}{r} \qquad (d)\;\cos \theta - \sin \theta=\frac{\sqrt 3}{r} $

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$(a)\;\sin \theta + \cos \theta=\frac{\sqrt 3+1}{r}$
answered Nov 7, 2013 by pady_1

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