Question

if ABCD is a cyclic quadrilateral than cos(180+A)+cos(180-B)+cos(180-C)-sin(90-D)

Answer 1

$\cos(180 + a ) = - \cos(a ) \\ \cos(180 - b ) = - \cos(b) \\ \cos(180 - c) = - \cos(c ) \\ \sin(90 - d) = \cos(d)$

Answer 2

Answer:

The answer is 0.

Step-by-step explanation:

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

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    ( Since, cos (180±x) = - cos x and sin (90 - x) = cos x )

= 0.