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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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2 (cos A + cos C)

2 (cos A + cos B)

2 (cos A + cos D)

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detailed solution

Correct option is D

Given that, ABCD is a cyclic quadrilateral. So, A+C=180∘⇒A=180∘−C⇒cos⁡A=cos⁡180∘−C=−cos⁡C⇒cos⁡A+cos⁡C=0---i Similarly, cos⁡B+cos⁡D=0----iiOn adding Eqs. (i) and (ii), we getcosA + cosB + cosC + cosD =0

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