Question

Cos(180'-A) + Cos (180+B)+ cos(180 +c) - Sim (90+D) ​

Answer 1

Hope It Will Help........

Answer 2

Answer: Given,

ABCD is a cyclic quadrilateral

We know that,

The sum of the opposite angles of a cyclic quadrilateral is 180°,

⇒ A + C = 180

⇒ A = 180 - C

Similarly,

B + D = 180

⇒ D = 180 - B

cos (180±x) = - cos x

sin (90 - x) = cos x

Thus,

cos(180+A)+cos(180-B)+cos(180-C)-sin(90-D)

= cos(180+A)+cos(D)+cos(A)-sin(90-D)

= - cos A + cos D + cos A - cos D    

= 0.