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Question

If A, B, C, D, are the angles of cyclic quadrilateral, then cosA + cosB + cosC + cosD =


A

2(cosA + cosC)

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B

2 (cosB + cosD)

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C

2 (cosA + cosD)

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D

0

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Solution

The correct option is D

0


Since A, B, C, D is the angles of cyclic quadrilateral

Sum of opposite angles of quadrilateral = 180

A + C = 180

B + D = 180

= cosA + cos B + cosC + cosD

= cosA + cosB + cos (180-A) + cos(180-B)

= cosA + cosB + (-cosA) + (-cosB)

= cosA + cosB - cosA - cosB

= 0


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