White hole region in Kruskal–Szekeres coordinates and geodesic We use Kruskal–Szekeres coordinates$(T,X)$ to describe black hole and white hole.
For $r>2GM:$
$$T = \left(\frac{r}{2GM}-1\right)^{1/2}e^{r/4GM}\sinh\left(\frac{t}{4GM}\right)\\$$
$$X = \left(\frac{r}{2GM}-1\right)^{1/2}e^{r/4GM}\cosh\left(\frac{t}{4GM}\right)$$
For $r<2GM:$
$$T = \left(1-\frac{r}{2GM}\right)^{1/2}e^{r/4GM}\cosh\left(\frac{t}{4GM}\right)$$
$$X = \left(1-\frac{r}{2GM}\right)^{1/2}e^{r/4GM}\sinh\left(\frac{t}{4GM}\right)$$
From Wiki, the graph of Kruskal–Szekeres coordinates can be divided into 4 regions. However, I cant get get both negative $T$ and $X$ from the above definition since $\cosh$ is always positive, then, how can we  obtain curves at the left bottom part of the graph?

Also, I am wondering can I obtain some formula for drawing the geodesic on the graph other than radial null geodesic?
 A: I think probably the best equation to start from are (with $G = 1$)
\begin{align}
-T^2 + X^2 &= \left( \frac{r}{2 M} - 1 \right) \exp(\frac{r}{2M})
\end{align}
Here you can see that negating one or both of $T$ and $X$ does not modify the above equation, making it true in all patches. To solve for a particular patch you have to choose if $T$ and $X$ are greater than or equal to $0$ in the patch you want when you take the square root.
The other equation is
$$
\frac{T+X}{T-X} = \exp(\frac{t}{2M}).
$$
This equation holds for both the right and the left patch. When it comes to the the top and bottom patch, $t$ does not actually cover those patches. From the above equation, you can see how if the horizon is $T = \pm X$, that corresponds to $t = \pm \infty$. That means that you can't continue $t$ the $|T| > |X|$ region.
However, if you make the left hand side negative, as in
$$
\frac{T+X}{T-X} = -\exp(\frac{t}{2M}).
$$
then now you can get the upper and lower regions, but not the right and left regions. If you solve for $T$ and $X$ in terms of $t$ and $r$ in the region you want, you can get curves in any region you want.
