if Δ1=xsinθcosθ−sinθ−x1cosθ1x and Δ2=xsin2θcos2θ−sin2θ−x1cos2θ1x  x≠0, then for all θ∈0,π2

We have Δ1=xsinθcosθ-sinθ-x1cosθ1x                       =x-x2-1+sinθxsinθ-cosθ+cosθsinθ+xcosθ                        =-x3           and  Δ2=xsin2θcos2θ-sin2θ-x1cos2θ1x                         =x-x2-1+sin2θxsin2θ-cos2θ+cos2θsin2θ+xcos2θ                         =-x3 ∴ Δ1+ Δ2 =-2x3