Math, asked by aashish1360, 1 year ago

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 abhi178
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 adityaroykapoor2684
0

Answer: to

Step-by-step explanation:

Attachments:
Similar questions