Question

if a b c d circle quadrilateral then
prove that

cos A + cos B + cos C + cos D =0​

Answer 1

Answer:

ABCD is a cyclic quadrilateral.

We know that opposite angles of cyclic quadrilateral are supplementary.

∴ A+C=π=180

⇒ A=π−C

⇒ cosA=cos(π−C)

⇒ cosA=cosπ.cosC+sinπ.sinC

⇒ cosA=(−1).cosC+0.sinC

⇒ cosA=−cosC

∴ cosA+cosC=0 ----- ( 1 )

∴ B+D=π=180

⇒ B=π−D

⇒ cosB=cos(π−D)

⇒ cosB=cosπ.cosD+sinπ.sinD

⇒ cosB=(−1).cosD+0.sinD

⇒ cosB=−cosD

∴ cosB+cosD=0 ---- ( 2 )

Adding ( 1 ) and ( 2 ) we get,

⇒ cosA+cosB+cosC+cosD=0

Answer 2

Answer:

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)

PS: In Euclidean geometry, a cyclic quadrilateral is a quadrilateral (polygon of four sides) inscribed in a circle such that all its four vertices lie on the circumference of the circle. The circle is called the circumscribed circle. All quadrilaterals do not have circumscribed circles. The word cyclic comes from the Greek word KUKλOξ (kuklos) which means ‘wheel’ or ‘circle’.

Step-by-step explanation:

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