Chapter 13, Trigonometric Ratios and Functions - 13.3 Evaluate Trigonometric Functions of Any Angle - 13.3 Exercises - Skill Practice - Page 871: 29

$\dfrac{\sqrt 3}{3}$

When $P$ is the period of $f(x)$ then we have $ f(x+p)=f(x)$ Here, we have $ \cot (-\dfrac{8 \pi}{3})=-\cot (-\dfrac{8 \pi}{3})$ But $\dfrac{8 \pi}{3}- 2\pi=\dfrac{8 \pi}{3}-\dfrac{6 \pi}{3}=\dfrac{2 \pi}{3}$ Thus, we have $-\cot(\dfrac{2 \pi}{3})=-\tan [-\dfrac{\pi}{6}]$ or, $=\dfrac{\sqrt 3}{3}$