Question

In any quadrilateral ABCD find the value of sin(A+B) +sin(C+D)

Answer 1

in any quadrilateral ABCD
A+B+C+D=360°
or A+B=360°-(C+D)
or sin(A+B)=sin[360°-(C+D)]
or sin(A+B)= -sin(C+D)
or sin(A+B)+sin(C+D)=0. answer

Answer 2

The value of $\bold\sin (A+B)+\sin (C+D) \text { is } 0}$

To find:

The value of \sin (A+B)+\sin (C+D)

Solution:

By using the trigonometric identities

We all know that,

With the help of the property of cyclic quadrilateral, that the sum of the angles in quadrilaterals is 360.

By applying the cyclic quadrilateral property to the given quadrilateral of having sides A, B, C, D  

A+B+C+D = 360.

A+B = 360- (C+D)

Applying sins on both sides, then the results we get,

$\sin (A+B)=\sin (360-(C+D))$

$\sin (A+B)=-\sin (C+D)$

$\sin (A+B)+\sin (C+D)=0$