If A, B, C, D are the angles of a cyclic quad | vedclass

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(D) Given that $ABCD$ is a cyclic quadrilateral.In a cyclic quadrilateral,the sum of opposite angles is $180^\circ$.Therefore,$A + C = 180^\circ$ and

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(D) Given that $ABCD$ is a cyclic quadrilateral.In a cyclic quadrilateral,the sum of opposite angles is $180^\circ$.Therefore,$A + C = 180^\circ$ and $B + D = 180^\circ$.From $A + C = 180^\circ$,we get $A = 180^\circ - C$.Taking cosine on both sides,$\cos A = \cos(180^\circ - C) = -\cos C$.This implies $\cos A + \cos C = 0$.Similarly,from $B + D = 180^\circ$,we get $B = 180^\circ - D$.Taking cosine on both sides,$\cos B = \cos(180^\circ - D) = -\cos D$.This implies $\cos B + \cos D = 0$.Adding these two equations,we get $\cos A + \cos B + \cos C + \cos D = 0 + 0 = 0$.

If $A, B, C, D$ are the angles of a cyclic quadrilateral,then $\cos A + \cos B + \cos C + \cos D = $

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