Question

prove that Cos A + cos b + cos C + cos D=0​

Answer 1

Pls complete the question...

Answer 2

Answer:

Step-by-step explanation:

As we know that...

In cyclic quadrilateral, the sum of opposite angles is 180°.

SO.. A+C= 180° and B+D= 180°

HENCE,

cosA+cosC = 2cos(A+C/2)cos(A-C/2)

cosA+cosC = 2cos(180°/2)cos(A-C/2)

cosA+cosC= 0 {as cos90°=0}

cosB+cosD = 2cos(B+D/2)cos(B-D/2)

cosB+cosD = 2cos90°cos(B-D/2)

cosB+cosD = 0

cosA+cosB+cosC+cosD=0

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