Answer:
[tex]\frac{3\sqrt{2}+2\sqrt{3}}{6}[/tex] or [tex]\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}[/tex]
Step-by-step explanation:
[tex]\sin\frac{3\pi}{4}-\tan\frac{5\pi}{6}\\\\\sin\frac{3\pi}{4}-\frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}}\\ \\\frac{\sqrt{2}}{2}-\frac{\frac{1}{2} }{-\frac{\sqrt{3}}{2}}\\ \\\frac{\sqrt{2}}{2}+\frac{1}{\sqrt{3}}\\ \\\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}\\ \\\frac{3\sqrt{2}}{6}+\frac{2\sqrt{3}}{6}\\ \\\frac{3\sqrt{2}+2\sqrt{3}}{6}[/tex]
