If f(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.