A) 0
B) -1
C) 1
D) Cannot be found
Correct Answer: A
Solution :
Since, ABCD is a cyclic Quadrilateral, therefore \[\angle A+\angle C=\angle B+\angle D={{180}^{o}}\] or \[\frac{A+C}{2}=\frac{B+D}{2}={{90}^{o}}\] \[\therefore \] \[\cos A+\cos B+\cos C+\cos D\] \[=2\cos \frac{A+C}{2}\cos \frac{A-C}{2}+2\cos \frac{B+D}{2}\cos \frac{B-C}{2}\] \[=2\cos {{90}^{o}}\cos \frac{A-C}{2}+2\cos {{90}^{o}}\cos \frac{B-C}{2}\] \[=0\]
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