Question

If ABCD is a cyclic quadrilateral prove that cosA+cosB+cosC+cosD=0 ..

Answer 1

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|

I WOULD SUGGEST YOU TO TAKE A LOOK AT THE CONCEPT OF ASSOCIATED ANGLES, WHICH IS IN THE SYLLABUS OF CLASS XI.

THANKS.

Answer 2

Answer:

Since, the in a cyclic quadrilateral,

The sum of opposite angles is 180°,

Given,

ABCD is quadrilateral,

⇒ ∠A + ∠C = 180°,

⇒ ∠A = 180° - ∠C, ------(1)

Similarly,

∠B = 180° - ∠D, --------(2)

Taking cosine on both sides of equation (1),

cos A = cos (180° - C)

⇒ cos A = -cos C            ( Since, cos ( 180 - x ) = - cos x )

Again, Taking cosine on both sides of equation (2),

cos B = cos (180° - D)

⇒ cos B = -cos D  

Thus, cosA + cosB + cosC + cosD

= cosA - cosA + cosC - cosC = 0

Hence, proved.....