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

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

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(A) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$. Thus,$A + C = 180^{\circ}$ and $B + D = 180^{\circ}$.Using trigonometric identities:$\cos(180^{\circ} + A) = -\cos A$$\cos(180^{\circ} - B) = -\cos B$$\cos(180^{\circ} - C) = -\cos C$$\sin(90^{\circ} - D) = \cos D$Substituting these into the expression:$-\cos A - \cos B - \cos C - \cos D = -(\cos A + \cos C) - (\cos B + \cos D)$Since $C = 180^{\circ} - A$,$\cos C = \cos(180^{\circ} - A) = -\cos A$,so $\cos A + \cos C = 0$.Similarly,$D = 180^{\circ} - B$,$\cos D = \cos(180^{\circ} - B) = -\cos B$,so $\cos B + \cos D = 0$.Therefore,the expression equals $-(0) - (0) = 0$.

If $A, B, C, D$ are the angles of a cyclic quadrilateral taken in order,then $\cos(180^{\circ} + A) + \cos(180^{\circ} - B) + \cos(180^{\circ} - C) - \sin(90^{\circ} - D) =$

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