Answer:
A
Step-by-step explanation:
[tex]\frac{11\pi }{6}[/tex] is in the fourth quadrant where tan < 0
The related acute angle = 2π - [tex]\frac{11\pi }{6}[/tex] = [tex]\frac{\pi }{6}[/tex]
Hence
tan( [tex]\frac{11\pi }{6}[/tex] )
= - tan([tex]\frac{\pi }{6}[/tex] )
= - [tex]\frac{1}{\sqrt{3} }[/tex]
= - [tex]\frac{1}{\sqrt{3} }[/tex] × [tex]\frac{\sqrt{3} }{\sqrt{3} }[/tex]
= - [tex]\frac{\sqrt{3} }{3}[/tex] → A
