Answer
The graphs of the curve and its tangent lines are below.
Work Step by Step
$$y=f(x)=\tan x\hspace{1cm}\{-\pi/2\le x\le \pi/2\}$$
The derivative of $f(x)$: $$f'(x)=\sec^2 x$$
The tangent lines to $f(x)$ are:
1) At $x=-\pi/3$:
$f(x)=\tan(-\pi/3)=-\sqrt3$
$f'(x)=\sec^2(-\pi/3)=\frac{1}{\cos^2(-\pi/3)}=\frac{1}{(-1/2)^2}=4$
$$y-(-\sqrt3)=4(x-(-\frac{\pi}{3}))$$ $$y+\sqrt3=4(x+\frac{\pi}{3})=4x+\frac{4\pi}{3}$$ $$y=4x+\frac{4\pi}{3}-\sqrt3=4x+\frac{4\pi-3\sqrt3}{3}$$
2) At $x=0: f(x)=\tan(0)=0$ and $f'(x)=\sec^2(0)=\frac{1}{\cos^20}=1$
$$(y-0)=1(x-0)$$ $$y=x$$
3) At $x=\pi/3$:
$f(x)=\tan(\pi/3)=\sqrt3$
$f'(x)=\sec^2(\pi/3)=\frac{1}{\cos^2(\pi/3)}=\frac{1}{(1/2)^2}=4$
$$y-\sqrt3=4(x-\frac{\pi}{3})=4x-\frac{4\pi}{3}$$ $$y=4x-\frac{4\pi}{3}+\sqrt3=4x+\frac{3\sqrt3-4\pi}{3}$$
The graphs of the curve and its tangent lines are below.