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.Since the sum of opposite angles in a cyclic quadrilateral is $180^\circ$,we have $A + C = 180^\circ$

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(D) Given that $ABCD$ is a cyclic quadrilateral.Since the sum of opposite angles in a cyclic quadrilateral is $180^\circ$,we have $A + C = 180^\circ$ and $B + D = 180^\circ$.From $A + C = 180^\circ$,we get $A = 180^\circ - C$.Therefore,$\cos A = \cos(180^\circ - C) = -\cos C$,which implies $\cos A + \cos C = 0$.Similarly,from $B + D = 180^\circ$,we get $B = 180^\circ - D$.Therefore,$\cos B = \cos(180^\circ - D) = -\cos D$,which implies $\cos B + \cos D = 0$.Adding these two results,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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