Question

if ABCD is cyclic quadrilateral then cos A + cos B is equal to A) 0 B) cos c + cos D c) -( cos c + cos D ) d) cos c - cos D​

Answer 1

Answer:

cos A + cos B is equal to A) 0

Step-by-step explanation:

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Answer 2

Answer:

∴ A+C=π=180

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⇒ 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

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⇒ 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