Answer
The graphs of the curve and its tangent lines are below.
Work Step by Step
$$y=f(x)=\sec x\hspace{1cm}\{-\pi/2\le x\le \pi/2\}$$
The derivative of $f(x)$: $$f'(x)=\sec x\tan x$$
The tangent lines to $f(x)$ are:
1) At $x=-\pi/3$:
$f(x)=\sec(-\pi/3)=\frac{1}{\cos(-\pi/3)}=\frac{1}{\frac{1}{2}}=2$
$f'(x)=\sec(-\pi/3)\tan(-\pi/3)=2\times(-\sqrt3)=-2\sqrt3$
$$y-2=-2\sqrt3(x+\frac{\pi}{3})=-2\sqrt3x-\frac{2\sqrt3\pi}{3}$$ $$y=-2\sqrt3x-\frac{2\sqrt3\pi}{3}+2=-2\sqrt3x+\frac{6-2\sqrt3\pi}{3}$$
2) At $x=\pi/4$:
$f(x)=\sec(\pi/4)=\frac{1}{\cos(\pi/4)}=\frac{1}{\frac{\sqrt2}{2}}=\sqrt2$
$f'(x)=\sec(\pi/4)\tan(\pi/4)=\sqrt2\times1=\sqrt2$
$$y-\sqrt2=\sqrt2(x-\frac{\pi}{4})=\sqrt2x-\frac{\sqrt2\pi}{4}$$ $$y=\sqrt2x-\frac{\sqrt2\pi}{4}+\sqrt2=\sqrt2x+\frac{4\sqrt2-\sqrt2\pi}{4}$$
The graphs of the curve and its tangent lines are below.