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IfABCD is a cyclic quadrilateral then what is $\sin A+\sin B-\sin C-\sin D$ equal to?
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We know in cyclic quadrilateral, ABCD $\mathrm{A}+\mathrm{C}=180^{\circ}, \mathrm{B}+\mathrm{D}=180^{\circ}$
$\therefore \mathrm{A}=180^{\circ}-\mathrm{C}, \mathrm{B}=180^{\circ}-\mathrm{D}$
$\sin \mathrm{A}+\sin \mathrm{B}-\sin \mathrm{C}-\sin \mathrm{D}$
$=\sin \left(180^{\circ}-\mathrm{C}\right)+\sin \left(180^{\circ}-\mathrm{D}\right)-\sin \mathrm{C}-\sin \mathrm{D}$
$=\sin \mathrm{C}+\sin \mathrm{D}-\sin \mathrm{C}-\sin \mathrm{D}=0$
$\therefore \mathrm{A}=180^{\circ}-\mathrm{C}, \mathrm{B}=180^{\circ}-\mathrm{D}$
$\sin \mathrm{A}+\sin \mathrm{B}-\sin \mathrm{C}-\sin \mathrm{D}$
$=\sin \left(180^{\circ}-\mathrm{C}\right)+\sin \left(180^{\circ}-\mathrm{D}\right)-\sin \mathrm{C}-\sin \mathrm{D}$
$=\sin \mathrm{C}+\sin \mathrm{D}-\sin \mathrm{C}-\sin \mathrm{D}=0$
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