Question

If abcd be the anvles of a cycle quadrilateral taken in order,prove that cos(180-a)+cos(180+b)+cos(180+c)-sin(90+d)=0

Answer 1

cos(180-A) = cos C as A+C= 190 degrees 
and cos(180+C) = -cosC 
Hence cos (180 - A) + cos (180 + B) + cos (180 + C) - sin (90 - D) = cos(180+B) - sin(90-D) 
sin(90-D) = sin(90-(180-B)) = sin(-(90-B)) = -sin (90-B) = -cosB where B is assumed acute angled as 90-B is considered positive 
hence cos(180+B) - sin(90-D) = -cosB - (-cosB) = 0

Answer 2

Answer:

Here,

ABCD is a cyclic quadrilateral

Since, the sum of the opposite angles of a cyclic quadrilateral is supplementary,

In quadrilateral ABCD,

A and C are opposite angles,

⇒ A + C = 180

⇒ A = 180 - C

Similarly, B and D are opposite angles,

B + D = 180

⇒ D = 180 - B

L.H.S.

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

= 0 = R.H.S.

Hence, proved.....