(D) Since the quadrilateral $ABCD$ is cyclic,the sum of opposite angles is $180^{\circ}$.Therefore,$A + C = 180^{\circ}$ and $B + D = 180^{\circ}$.Thi

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(D) Since the quadrilateral $ABCD$ is cyclic,the sum of opposite angles is $180^{\circ}$.Therefore,$A + C = 180^{\circ}$ and $B + D = 180^{\circ}$.This implies $A = 180^{\circ} - C$ and $B = 180^{\circ} - D$.Using the property $\cos(180^{\circ} - \theta) = -\cos \theta$,we get:$\cos A = \cos(180^{\circ} - C) = -\cos C \implies \cos A + \cos C = 0$.$\cos B = \cos(180^{\circ} - D) = -\cos D \implies \cos B + \cos D = 0$.Adding these,we get $\cos A + \cos B + \cos C + \cos D = 0 + 0 = 0$.

Class 11 Mathematics Trigonometrical Equations Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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If $A, B, C, D$ are the angles of a cyclic quadrilateral taken in order,then $\cos A + \cos B + \cos C + \cos D =$

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$-1$
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$1$
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$\frac{1}{2}$
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$0$

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Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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