## Calculus (3rd Edition)

The equation of the tangent line is given by $$y=\frac{4}{3} \theta+\frac{\sqrt{3}}{3}-\frac{ 2\pi}{9} .$$
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 $$y=\frac{4}{3} \theta+\frac{\sqrt{3}}{3}-\frac{ 2\pi}{9} .$$