${\cos }^2{105}^{\circ }-{\sin }^2{105}^{\circ }=\dots ={\sin }^2{15}^{\circ }-{\cos }^2{15}^{\circ }=\frac{1}{8+4\sqrt{3}}-\frac{1}{8-4\sqrt{3}}=\frac{8-4\sqrt{3}-\left(8+4\sqrt{3}\right)}{\left(8+4\sqrt{3}\right)\left(8-4\sqrt{3}\right)}=\frac{8-4\sqrt{3}-8-4\sqrt{3}}{64-48}=\frac{-8\sqrt{3}}{16}=-\frac{\sqrt{3}}{2}$