the region is described in polar coordinates $$D= (r,\theta): \begin{cases} 0\leq r \leq 2\sec(\theta) \\ -\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4} \end{cases}$$ and the exercise asks to draw it on the $xy$ plane.
I attach a picture of how I drew it
the truth i'm not sure if that's okay.
On
$r = 2\sec \theta\\ r\cos \theta = 2\\ x = r\cos \theta\\ x = 2$
This is a vertical line.
$\theta = \pm \frac {\pi}{4}$ gives the endpoints of the line.
$(2,-2), (2,2)$
But we have an inequality. $r \le 2\sec \theta$
This means that we include everything between our line segment and the origin.
Or we have a isosceles right triangle.
For each angle $\theta$ between $-\pi/4$ and $\pi/4$, the value of $r$ lies between $0$ and $2\sec \theta$. That's literally what the inequalities state. So if we pick a representative angle $\theta$ and draw a line from the origin with this angle, then stop when the line has length $2 \sec \theta$, where does it stop? Call this endpoint $(x,y)$.
Well, if we drop a perpendicular to the $x$-axis from the endpoint, then we form a right triangle whose hypotenuse has length $r = 2 \sec \theta$, and $$\cos \theta = \frac{x}{r} = \frac{x}{2\sec \theta}, \\ \sin \theta = \frac{y}{r} = \frac{y}{2 \sec \theta}.$$ Therefore, the $x$-value of this coordinate is $$x = 2 \sec \theta \cos \theta = 2,$$ and the $y$-value of this coordinate is $$2 \sec \theta \sin \theta = 2 \tan \theta.$$ So $$(x,y) = (2, 2\tan \theta)$$ for each $\theta \in [-\pi/4, \pi/4]$. Notice that because the $x$-value is always $2$, and $\tan (-\pi/4) = -1$ and $\tan \pi/4 = 1$, the set of all endpoints of these radial line segments is the line segment that connects $(2,-2)$ and $(2,2)$. So the region that is enclosed by the given inequalities is a triangle with vertices $(0,0)$, $(2,-2)$, and $(2,2)$. This region is swept by a radial line segment from the origin with angle ranging from $-\pi/4$ to $\pi/4$, and endpoint on the line $x = 2$.