Correct Answer - a
ABCD is concyclic quadrilateral
`angleA+angleC=angleB+angleD=180^(@)`
`therefore angle A=180^(@)-angleC`
`cosA=cos(180^(@)-C)rArr-cosC`
Similarly
`Cos B=-cosD`
`rArrcosA+cosB+cosC+cosD`
`rArrcosA+cosB-cosA-cosB=0`
Alternate
put, A=B=C=D=`90^(@)`
`=cosA+cosB+cosC+cosD`
`=cos90^(@)+cos90^(@)+cos90^(@)+cos90^(@)`
`=0+0+0+0=0`
