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If any quadrilateral ABCD, prove that`"sin"(A+B)+sin(C+D)=0``"cos"(A+B)=cos(C+D)`

Answer» In quadrilateral `ABCD, A+B+C+D=2pi`.
(a) `sin(A+B)+sin(C+D)=sin(A+B)+sin(2pi-(A+B))`
`=sin(A+B)-sin(A+B)=0`
(b) `cos(A+B)=cos(C+D)`
`=cos(2pi-(C+D))=cos(C+D)`


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