Question

What is the answer for this question?

Answer 1

Answer:

A

Step-by-step explanation:

$\frac{11\pi }{6}$ is in the fourth quadrant where tan < 0

The related acute angle = 2π - $\frac{11\pi }{6}$ = $\frac{\pi }{6}$

Hence

tan( $\frac{11\pi }{6}$ )

= - tan($\frac{\pi }{6}$ )

= - $\frac{1}{\sqrt{3} }$

= - $\frac{1}{\sqrt{3} }$ × $\frac{\sqrt{3} }{\sqrt{3} }$

= - $\frac{\sqrt{3} }{3}$ → A