lf A,B,C and D are the angles of a cyclic quadrilateral, then cos A + cos B + cos C + cos D is equal to
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a
2 (cos A + cos C)
b
2 (cos A + cos B)
c
2 (cos A + cos D)
d
0
answer is D.
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Detailed Solution
Given that, ABCD is a cyclic quadrilateral. So, A+C=180∘⇒A=180∘−C⇒cosA=cos180∘−C=−cosC⇒cosA+cosC=0---i Similarly, cosB+cosD=0----iiOn adding Eqs. (i) and (ii), we getcosA + cosB + cosC + cosD =0