If A,B,C and D are the angles of cyclic quadrilateral, prove that:
i) `cosA + cosB+cosC+cosD`= 0
ii) `cos(180^(@)-A)+cos(180^(@)+B)+cos(180^(@)+C)-sin(90^(@)+D)`= 0

Explanation

A,B,C and D are the angles of a cyclic quadrilateral.
Therefore, A+C=`180^(@)`
`B+D=180^(@)`……………………(1)
i) LHS `=cosA+cosB+cosC+cosD`
`=cosA+cosB+cos(180^(@)-A) + cos(180^(@)-B)` from eq. (1)
`=cosA+cosB-cosA_cosB=0`
= RHS Hence proved.
ii) LHS `=cos(180^(@)-A)+cos(180^(@)+B) + cos(180^(@)+C)-sin(90^(@)+D)`
`=-cosA-cosB-cosC-cosD`
`=-cosA+cosB+cosC+cosD`
=0 (form part (i))
=RHS Hence proved.