If θ is eliminated from the equations x=acos⁡(θ−α) and y=bcos⁡(θ−β), then x2a2+y2b2−2xyabcos⁡(α−β) is equal to

(α−β)=(θ−β)−(θ−α)⇒ cos⁡(α−β)=cos⁡(θ−β)cos⁡(θ−α)+sin⁡(θ−β)sin⁡(θ−α)=yb×xa+1−x2a21−y2b2⇒     xyab−cos⁡(α−β)2=1−x2a21−y2b2 or      x2y2a2b2+cos2⁡(α−β)−2xyabcos⁡(α−β)    =1−y2b2−x2a2+x2y2a2b2 or  x2a2+y2b2−2xyabcos⁡(α−β)=sin2⁡(α−β)