In a cyclic quadrilateral ABCD, the value of
sinA/2+sinB/2-cosC/2-cosD/2 is
(1) 0 (2) 1
(3) 2 (4) 3
Answers
Answered by
1
answer : option (1) 0
explanation : ABCD is a cyclic quadrilateral.
so, A + C = B + D = 180°
or, A/2 + C/2 = B/2 + D/2 = 90°
C/2 = 90° - A/2 ........(1)
and D/2 = 90° - B/2 .........(2)
now, sinA/2 + sinB/2 - cosC/2 - cosD/2
from equations (1) and (2),
= sinA/2 + sinB/2 - cos(90° - A/2) - cos(90° - B/2)
we know, cos(90° - x) = sinx
so, cos(90° - A/2) = sinA/2
cos(90° - B/2) = sinB/2
= sinA/2 + sinB/2 - sinA/2 - sinB/2
= 0
hence, sinA/2 + sinB/2 - cosC/2 - cosD/2 = 0
Answered by
0
Answer: to
Step-by-step explanation:
Attachments:

Similar questions