Answer
The equation of the tangent line is given by
$$ y=\frac{4}{3} \theta+\frac{\sqrt{3}}{3}-\frac{ 2\pi}{9} .$$
Work Step by Step
Since $ y=\tan \theta $, then $ y'=\sec^2 \theta $ and hence the slope at $ x=\frac{\pi}{6}$ is $ m=\sec^2 \frac{\pi}{6}=\frac{4}{3}.$ Now, the equation of the tangent line at $ x=\frac{\pi}{6}$ is given by
$$ y=\frac{4}{3} \theta+c.$$
Since the curve and line coincide at $\theta=\frac{\pi}{6}$, then
$$\tan \frac{\pi}{6}=\frac{4}{3}\frac{\pi}{6}+c \Longrightarrow c= \frac{1}{\sqrt{3}}-\frac{ 2\pi}{9} .$$
Hence the equation of the tangent line is given by
$$ y=\frac{4}{3} \theta+\frac{\sqrt{3}}{3}-\frac{ 2\pi}{9} .$$
