if ABCD is a cyclic quadrilateral, then find the value of cosA+cosB+cosC+cosD.
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As A, B, C, D are the vertices of CyQuad ABCD,
A+C=180° and B+D=180°
By the rules of *associated angles*, we can write,
CosC=Cos(2x90-A)= -CosA.....(1)
We can apply the same process on CosB or CosD......
Then we shall get,
CosB=-CosD................................ (2)
So, in L. H. S.
CosA+CosB+CosC+CosD
=CosA-CosD-CosA+CosD........ From 1&2
=0
Hence, L.H.S=R.H.S., |PROVED|
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