I am wondering what it looks like the maximal torus of the Borel subgroup of $SO(2n)(\mathbb{R})$.
I guess that $(2 \times 2)$ matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos \theta \end{pmatrix}$ occurs in the diagonal as well $t \in \mathbb{R}^{\times}$.
Thank you for your comment in advance.
There is no Borel subgroup of $SO(n)$ as it is compact. You can say $SO(n)$ is the only parabolic subgroup of itself but it is still not Borel.
To have Borel subgroups a (semisimple) group must be what is called quasi-split. The only real forms of the complex special orthogonal group that are quasi-split are those of the form $SO(n,n), SO(n,n+1), SO(n,n+2)$. The first two being the split case in even and odd dimensions respectively and the last being the non-split case in even dimensions (no non-split, quasi-split case in odd dimensions)
Over the complex numbers, all tori are contained in a Borel subgroup and all are conjugate so nothing to say really there but it is important to note there is more than one and this is true in the real forms as well. Indeed there will be one for every Borel subgroup opposite to the starting one which is an affine space of dimension equal to that of the unipotent radical.
In the split case I think they should be split tori (i.e. diagonalisable). In the quasi-split case I'm less sure. From here we can see that if the group was adjoint or simply connected then the tori would have to be quasi-split ones but $SO(n,n+2)$ is neither (indeed it's not even connected but I assume we'd implicitly pass to the connected component of the identity)