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If $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are the angles of a cyclic quadrilateral taken in order, then $\cos A+\cos B+\operatorname{Cos} C+\operatorname{Cos} D=$
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Since the quadrilateral $\mathrm{ABCD}$ is cyclic, we have
$\mathrm{A}+\mathrm{C}=180^{\circ}$ and $\mathrm{B}+\mathrm{D}=180^{\circ}$
$\therefore \cos \mathrm{A}=\cos \left(180^{\circ}-\mathrm{C}\right)=-\cos \mathrm{C}$
$\cos \mathrm{B}=\cos \left(180^{\circ}-\mathrm{D}\right) \quad=-\cos \mathrm{D}$
$\therefore \cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}+\cos \mathrm{D}=0$
$\mathrm{A}+\mathrm{C}=180^{\circ}$ and $\mathrm{B}+\mathrm{D}=180^{\circ}$
$\therefore \cos \mathrm{A}=\cos \left(180^{\circ}-\mathrm{C}\right)=-\cos \mathrm{C}$
$\cos \mathrm{B}=\cos \left(180^{\circ}-\mathrm{D}\right) \quad=-\cos \mathrm{D}$
$\therefore \cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}+\cos \mathrm{D}=0$
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