Verified answer

[tex]\displaystyle\bf\\Cos180^\circ\Big(Sin135^\circ+tg60^\circ-Cos135^\circ\Big)^{2} =\\\\\\=-1\cdot\Big[Sin(90^\circ+45^\circ)+\sqrt{3} -Cos(90^\circ+45^\circ)\Big]^{2} =\\\\\\=-\Big(Cos45^\circ+\sqrt{3}+Sin45^\circ\Big)^{2} =-\Big(\frac{\sqrt{2} }{2} +\sqrt{3}+\frac{\sqrt{2} }{2} \Big)^{2}=\\\\\\=-\Big(\sqrt{2} +\sqrt{3} \Big)^{2}=-\Big(2+2\sqrt{6} +3\Big)=-(5+2\sqrt{6} )[/tex]