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