I feel experts might be able to answer this question immediately.
Let $G$ be a connected $\mathbb Q$simple and $\mathbb Q$isotropic algebraic group.
Let $S$ be a maximal $\mathbb Q$split torus of $G$ and
let $T\supset S$ be a maximal torus defined over $\mathbb Q$. Let $T_a$ be the maximal anisotropic subtorus of $T$. Is $T$ a direct product of $S$ and $T_a$?

$\begingroup$ Thanks for the examples of user89334 and nfdc23. A further question is whether it is true that $T(\mathbb Q)=S(\mathbb Q) T_a(\mathbb Q)$. $\endgroup$ – ronggang Apr 17 '16 at 2:27
The Weilrestriction construction suggested by user89334 is $\mathbf{Q}$simple but not absolutely simple. To give absolutely simple examples, consider $G = {\rm{SL}}_n(D)$ for a central division algebra $D$ over $\mathbf{Q}$ with dimension $d^2>1$.
A maximal split torus in $G$ is given by the diagonal torus $S$ in the evident subgroup ${\rm{SL}}_n \subset G$. The centralizer $Z_G(S)$ is equal to $$\{(d_1,\dots,d_n) \in \underline{D}^{\times}\,\,\prod {\rm{Nrd}}(d_j) = 1\}$$ where $\underline{D}^{\times} := {\rm{GL}}_1(D)$, so $\mathscr{D}(Z_G(S))={\rm{SL}}_1(D)^n$. This derived group has center containing $S[d]$, and every maximal torus $T \subset Z_G(S)$ is uniquely an isogenous product $S \cdot T_0$ for a (necessarily anisotropic) maximal torus $T_0 \subset \mathscr{D}(Z_G(S))$. Such a maximal $T_0$ is the maximal anisotropic subtorus of $T$ and must contain the central $S[d]$ in the derived group due to its maximality, so $T$ is never a direct product of $S$ and $T_0$ since $d > 1$. Passing to (central) isogenous quotients of $G$ never makes that central overlap entirely go away (since $S[d]$ is not cyclic), so the phenomenon persists for all members of the (central) isogeny class.
No. Here is an easy to understand example to keep in mind.
Consider $\mathbb{C}^*$. It contains both the group $S^1$ and the group $\mathbb{R}^*$, but it is not their direct product (their intersection is $\{1,1\}$). Both groups are real tori, the compact one is anisotropic and the noncompact is split.
To see this example in a simple group, just consider the diagonal subgroup of $\text{SL}_2(\mathbb{C})$.
Formally, the algebraic group you want too look at here is the extension of scalars of $\text{SL}_2$ from $\mathbb{R}$ to $\mathbb{C}$. Let me call it $\mathbf{G}$. $\mathbf{G}$ happens to be isomorphic as a $\mathbb{C}$group to $\text{SL}_2\times \text{SL}_2$ which is not simple, but it has a real structure for which it is $\mathbb{R}$simple. In fact, since you started with an isotropic $\mathbb{Q}$simple group, the resulting group will be so too. $\mathbf{G}$ will contain the 2dim maximal $\mathbb{Q}$torus $T$ which is the extension of scalars of the (1dim) diagonal subgroup of $\text{SL}_2$. $T$ will contain $T_a\simeq \text{SO}_2$ and $S\simeq G_m$, but will not be their direct product.
Edit: this is an edit made in order to emphasize the $\mathbb{Q}$ structure in the above example and to address the question asked in a comment to the original question.
Let me redo the extension of scalars on $\text{SL}_2$, but this time from $\mathbb{Q}$ to its quadratic extension $\mathbb{Q}[i]$. This gives a $\mathbb{Q}$structure on $\mathbf{G}$ such that $\mathbf{G}(\mathbb{Q})=\text{SL}_2(\mathbb{Q}[i])$. The torus $T$ becomes $\mathbb{Q}$isomorphic to the extension of scalars of $\mathbf{G}_m$ from $\mathbb{Q}$ to $\mathbb{Q}[i]$. In particular, $T(\mathbb{Q})\simeq \mathbb{Q}[i]^*$ and under this isomorphism, $S(\mathbb{Q})\simeq \mathbb{Q}^*$ and $T_a(\mathbb{Q})=\{a+ib\mid a,b\in \mathbb{Q},~a^2+b^2=1\}$. Check that the inclusion $S(\mathbb{Q})T_a(\mathbb{Q})<T(\mathbb{Q})$ is not an equation (e.g $1+i$ is in the RHS but not in the LHS).
If you wish for a higher level explanation, note that the difference $T(\mathbb{Q})/S(\mathbb{Q})T_a(\mathbb{Q})$ could be measured by the first Galois cohomology group of the norm torus, a nice discussion of which you could find in p. 73 of the excellent book "Algebraic Groups and Number Theory" by Platonov and Rapinchuk.
Let me conclude that the example I gave above is $\mathbb{Q}$simple, as asked, but not absolutely simple: as a $\mathbb{C}$group (in fact, as a $\mathbb{Q}[i]$group) it is not simple. If you wish for an absolutely simple example you can consult the other excellent answers here, or merely find a $\mathbb{Q}$embedding of our group $\mathbf{G}$ in $\text{SL}_n$ for some $n$ ($n=4$ will do).

7$\begingroup$ The statement "The groups happens to be ${\rm{SL}}_2 \times {\rm{SL}}_2$" is potentially confusing: that is as a $\mathbf{C}$group, not as an $\mathbf{R}$group (so it may create a mistaken picture for the intended $\mathbf{Q}$group). $\endgroup$ – nfdc23 Apr 16 '16 at 22:35

$\begingroup$ Notation fixed, definitions clarified  Thank you. $\endgroup$ – Uri Bader Apr 17 '16 at 18:09

$\begingroup$ I made further edits in order to emphasize the $\mathbb{Q}$structure in the above example and to address the question asked in a comment to the original question. $\endgroup$ – Uri Bader Apr 18 '16 at 6:29
As the example of nfdc23 shows, the answer is generally no. But maybe it helps to think about this question in a somewhat wider context, where the notions of split and anisotropic tori arise: the study of a connected reductive algebraic group defined over an arbitrary field $k$ (as in the 1965 paper by Borel and Tits). In the structure theory of such groups, it quickly becomes clear that the nature of $k$anisotropic groups depends heavily on $k$ (and is not understood for many familiar fields). Leaving that aside, Borel and Tits got a lot of unified information about the structure of a $k$isotropic group. Modulo the knowledge of $k$anisotropic groups, this leads ultimately to the Tits classification method. Of course, the special case $k=\mathbb{Q}$ is part of this story, but the main ideas are developed for all $k$.
Note especially that the question raised here never gets answered explicitly in the structure theory. Indeed, the maximal $k$anisotropic subtorus $T_a$ here is mentioned but does not play an important role. The key players include: a (nontrivial!) maximal $k$split torus $S$ (unique up to $k$conjugacy), along with its (reductive) centralizer in $G$ (which of course contains $T_a$), a minimal $k$parabolic subgroup containing $S$, and various data about the associated root systems and Weyl groups.
What the general theory reveals is the existence of an almostdirect product $ T=T_a\, S$: see for example Borel's 8.15 in GTM 126. But toward the end of their respective textbooks, Borel (in his expanded second edition) and Springer (in his later framework of $F$groups) develop a lot of finer detail about classical groups somewhat in the spirit of the answer by nfdc23.
One extreme, however, is the case of a quasisplit group $G$, in which a minimal $k$parabolic subgroup is a Borel subgroup (and which is the only type possible for finite or some other special fields).

$\begingroup$ In what sense are quasisplit groups an extreme case? Even here the difficulty of an only almostdirect product can arise. (Of course literally any difficulty that can arise for tori can arise for quasisplit groups, since tori are quasisplit!) I would think that split groups would be an (excessively) extreme case. $\endgroup$ – LSpice Oct 30 '18 at 21:14