A) \[1\]
B) \[2\cos A\]
C) \[-1\]
D) (0)\[0\]
Correct Answer: D
Solution :
\[\because \] As the opposite angles of a cyclic quadrilateral are supplementary. \[\therefore \]\[A+B=C+D=180{}^\circ \] \[\therefore \] \[\cos A=\cos \,\,(180{}^\circ -C)=-\cos C\] \[\cos B=\cos \,\,(180{}^\circ -D)=-\cos D\] \[\therefore \]\[\cos A+\cos B+\cos C+\cos D\] \[=\cos A+\cos B-\cos A-\cos B=0\]
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