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 $B + D = 180^{\circ}$.We need to evaluat

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$0$

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(A) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.Therefore,$A + C = 180^{\circ}$ and $B + D = 180^{\circ}$.We need to evaluate $\cos A + \cos B + \cos C + \cos D$.This can be written as $(\cos A + \cos C) + (\cos B + \cos D)$.Using the sum-to-product formula $\cos x + \cos y = 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}$:$\cos A + \cos C = 2 \cos \frac{A+C}{2} \cos \frac{A-C}{2} = 2 \cos 90^{\circ} \cos \frac{A-C}{2} = 2(0) \cos \frac{A-C}{2} = 0$.Similarly,$\cos B + \cos D = 2 \cos \frac{B+D}{2} \cos \frac{B-D}{2} = 2 \cos 90^{\circ} \cos \frac{B-D}{2} = 2(0) \cos \frac{B-D}{2} = 0$.Thus,$\cos A + \cos B + \cos C + \cos D = 0 + 0 = 0$.

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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