## if A,B,C,Dare angles of a cyclic quadrilateral .prove that cosA+cosB+cosC+cosD=0​

Question

if A,B,C,Dare angles of a cyclic quadrilateral .prove that cosA+cosB+cosC+cosD=0​

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3 weeks 2021-09-04T18:24:43+00:00 1 Answer 0 views 0

Step-by-step explanation:

If A,B, C,D are cylic quadrilateral then prove that cosA+cosB+cosC+cosD=0?

Assume that the given cyclic quadrilateral ABCD is convex. Then we know from one of its many properties that, the opposite angles of ABCD are supplementary, that is

A + C = π = 180°……………….………………….(1)

B + D = π = 180°……………….………………….(2)

Transposing C from left to right in eq. (1),

A = π – C

Taking cosines on both sides,

cos A = cos (π – C) = cos π . cos C + sin π . sin C = -1 .cos C + 0.sin C = – cos C

Or, cos A + cos C = 0………………………..……(3)

Similarly it can be shown from eq.(2) that

cos B + cos D = 0……………………..……………(4)

Adding eq.(3) and eq.(4), we get

cos A + cos B + cos C + cos D = 0 (Proved)