Question

19. IC A, B, C, D be the angles of a cyclic quadrilateral, show that
COSA + cosB + cosC + cosD = 0.
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Answer 1

Answer:

Given ABCD is cyclic

A+C=180  

B+D=180  

cosA=cos(180 −C)=−cosC...(1)

cosB=cos(180 −D)=−cosD...(2)

Adding eqs (1) and (2) we get

cosA+cosB=−cosC−cosD

=cosA+cosB+cosC+cosD=0

Hence proved

Answer 2

Step-by-step explanation:

A+C=180°

B+D=180°

cos A=cos(180°-C)=-cos C

similarly, cos B=-cos D

therefore, cos A+cos c=0

and cos B+cos D=0

cos A+cos C+ cos B+cos D=0