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