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

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

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(A) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.Therefore,$A + C = 180^{\circ}$ and $B + D = 180^{\circ}$.This implies $C = 180^{\circ} - A$ and $D = 180^{\circ} - B$.Now,consider the expression $\cos A + \cos B + \cos C + \cos D$.Substituting the values of $C$ and $D$:$\cos A + \cos B + \cos(180^{\circ} - A) + \cos(180^{\circ} - B)$Since $\cos(180^{\circ} - 	heta) = -\cos 	heta$,we have:$\cos A + \cos B - \cos A - \cos B = 0$.

If $A, B, C, D$ are angles of a cyclic quadrilateral,then $\cos A + \cos B + \cos C + \cos D$ is equal to

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