If ϕ(α,β)=cos⁡α−sin⁡α1sin⁡αcos⁡α1cos⁡(α+β)−sin⁡(α+β)1 then

cos⁡α−sin⁡α1sin⁡αcos⁡α1cos⁡(α+β)−sin⁡(α+β)1=cos⁡α−sin⁡α1sin⁡αcos⁡α1001+sin⁡β−cos⁡β[Applying R3→R3−R1(cos⁡β)+R2(sin⁡β) =(1+sin⁡β−cos⁡β)cos2⁡α+sin2⁡α =1+sin⁡β−cos⁡β which is independent of α.