Verified answer

[tex]\displaystyle\bf\\3tg\frac{\pi }{4} -Sin^{2} 60^\circ+Cos^{2} \frac{\pi }{6} -Ctg^{2} 30^\circ=\\\\\\=3\cdot 1-\Big(\frac{\sqrt{3} }{2} \Big)^{2}+\Big(\frac{1}{2} \Big)^{2} -\Big(\sqrt{3} \Big)^{2}= 3-\frac{3}{4} +\frac{1}{4} -3=-0,5[/tex]