If `A ,B ,C ,D` are angles of a cyclic quadrilateral, then prove that `cos A+cos B+cos C+cos D=0`

Explanation

LHS

`=(sinx-cosx+1)/(sinx+cosx-1)`

`=[(sinx-(cosx-1))]/(sinx+(cosx-1))*[(sinx+(cosx-1))]/(sinx+(cosx-1)`

`=(sin^2x-(cosx-1)^2)/(sinx+(cosx-1))^2`

`=(sin^2x-(cos^2x+1-2cosx))/(sin^2x+cos^2x+1-2cosx+2sinxcosx-2sinx)`

`=(1-cos^2x-cos^2x-1+2cosx)/(2+2sinxcosx-2cosx-2sinx)`

`=(2cos(1-cosx))/(2(1-sinx)(1-cosx))`

`=cosx/(1-sinx)`

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