Ответ:

[tex]-\frac{1+\sqrt{3} }{2}[/tex]

Объяснение:

[tex]\displaystyle sin\bigg(270^\circ-\frac{\pi }{6} \bigg)-cos\bigg(\pi +\frac{\pi }{6} \bigg)=sin(270^\circ-30^\circ)-cos(180^\circ+30^\circ)=sin270^\circ-sin30^\circ-cos180^\circ-cos30^\circ=-\not1-\frac{1}{2} +\not1-\frac{\sqrt{3} }{2} =-\frac{1+\sqrt{3} }{2}[/tex]