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In a triangle ABC, if $\cos (A - C) \cos B + \cos 2 B = 0,$ then $a^{2}, b^{2}, c^{2}$ are in

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Correct option is C

$\cos (A - C) \cos B = - \cos 2 B$

$\Rightarrow \cos (A - C) \cos (A + C) = \cos 2 B$

$\Rightarrow \cos^{2} A - \sin^{2} C = 1 - 2 \sin^{2} B$

$\Rightarrow 1 - \sin^{2} A - \sin^{2} C = \sin^{2} B$

$\Rightarrow \sin^{2} A + \sin^{2} C = 2 \sin^{2} B$

$\Rightarrow a^{2} + c^{2} = 2 b^{2}$ (by sine rule)

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In a triangle ABC, if cos (A-C) cosB+cos2B=0,then a2,b2,c2 are in