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

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(A) In a cyclic quadrilateral $ABCD$,the sum of opposite angles is $180^\circ$.Therefore,$A + C = 180^\circ$ and $B + D = 180^\circ$.From $A + C = 180^\circ$,we have $A = 180^\circ - C$.Taking cosine on both sides,$\cos A = \cos(180^\circ - C) = -\cos C$,which implies $\cos A + \cos C = 0$.Similarly,from $B + D = 180^\circ$,we have $B = 180^\circ - D$.Taking cosine on both sides,$\cos B = \cos(180^\circ - D) = -\cos D$,which implies $\cos B + \cos D = 0$.Now,consider the expression $\cos A - \cos B + \cos C - \cos D$.Rearranging the terms,we get $(\cos A + \cos C) - (\cos B + \cos D)$.Substituting the values derived above,we get $0 - 0 = 0$.

If $ABCD$ is a cyclic quadrilateral,then the value of $\cos A - \cos B + \cos C - \cos D = $

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