In great generality a Lie group mod its maximal compact subgroup is contractible (for example this is true for all connected Lie groups). Whenever this is true then the Lie group $ D $ is diffeomorphic to a cartesian product of its maximal compact $ K $ with a contractible piece $ D/K $. So (again assuming $ G $ is connected or semisimple or some other sufficient condition to guarantee $ D/K $ contractible) if $ D $ is dimension $ 2n $ and the maximal compact subgroup is dimension $ n $ then we have a diffeomorphism
$$
D \cong K \times (D/K)
$$
where $ D/K $ is a contractible piece of dimension equal to $ K $. So $ D $ is diffeomorphic to a trivial $ \dim(K) $ dimensional real vector bundle over $ K $. On the other hand, since $ K $ is a Lie group it must be parallelizable. That is, the tangent bundle of $ K $ is a trivial $ \dim(K) $ dimensional real vector bundle over $ K $. In other words, we have that the Lie group $ D $ is diffeomorphic to the tangent bundle of its maximal compact subgroup,
$$
D \cong T(K).
$$
A particular case of this is when $ G $ is a linear algebraic group whose real points $ G_\mathbb{R} $ are compact. A compact group is always the maximal compact subgroup of its complexification (this follows from the fact that for subgroups of a complex linear algebraic group maximal compact is equivalent to compact plus Zariski dense). In other words, $ G_\mathbb{R} $ is the ($ n $ dimensional) maximal compact subgroup of the ( $ 2n $ real dimensional) group $ G_\mathbb{C} $. So, taking $ D=G_\mathbb{C} $ and $ K=G_\mathbb{R} $ in the argument above, we have that the complex points are diffeomorphic to the tangent bundle of the real points,
$$
G_\mathbb{C} \cong T(G_{\mathbb{R}}).
$$
Again this argument requires that the real points are compact. For example $ G_\mathbb{R} $ compact implies $ G_\mathbb{C} $ reductive group and so by Iwasawa decomposition for reductive groups we have that $ G_\mathbb{C} $ mod its maximal compact is contractible.
Let $ G $ be a linear algebraic group and $ H $ a linear algebraic subgroup. Suppose the real points $ G_\mathbb{R}$, $H_\mathbb{R} $ are compact. Consider the manifold of complex points
$$
M_\mathbb{C}=G_\mathbb{C}/H_\mathbb{C}.
$$
Then is the tangent bundle of
$$
M_\mathbb{R}=G_\mathbb{R}/H_\mathbb{R}
$$
diffeomorphic to $ M_\mathbb{C} $?
$\DeclareMathOperator\SO{SO}$Motivation:
$$
\SO_{n+1}(\mathbb{C})/\SO_{n}(\mathbb{C})
$$
is diffeomorphic to the tangent bundle of the $ n $ sphere
$$
S^n= \SO_{n+1}(\mathbb{R})/\SO_{n}(\mathbb{R}).
$$
More Motivation: When $ H $ is trivial the result follows from the discussion above.
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SU{SU}\DeclareMathOperator\SO{SO}$Comment on the non-compact case: Without the compactness assumption this is false. For example we can take $ H $ trivial and $ G=\SL_n $ (and $ n\geq 2 $ ). Then $ G_\mathbb{C}=\SL_n(\mathbb{C}) $ is homotopy equivalent to $ \SU_n $ while the tangent bundle of $ G_\mathbb{R}=\SL_n(\mathbb{R}) $ is homotopy equivalent to $ \SL_n(\mathbb{R}) $ which is homotopy equivalent to $ \SO_n(\mathbb{R}) $. But $ \SU_n $ and $ \SO_n(\mathbb{R}) $ are not homotopy equivalent. Thus $ \SL_n(\mathbb{C}) $ and the tangent bundle $ T(\SL_n(\mathbb{R})) $ are not homotopy equivalent so they certainly are not diffeomorphic.
Possible counter examples to consider next:
The Steifel manifold $ O_4/O_2 $ or the Grassmanian $ O_4/(O_2 \times O_2) $. What do the tangent bundles look like? Are they diffeomorphic to the complex points?
Best Answer
This fact about the tangent space being the complexification is known to be true if $ G_\mathbb{R}/H_\mathbb{R} $ is a compact symmetric space. In other words, if $ H_\mathbb{R} $ is the fixed points of an involution of $ G_\mathbb{R} $. This includes, for example, all the spheres written as $$ S^n \cong SO_{n+1}(\mathbb{R})/SO_n(\mathbb{R}) $$ This is taken directly from https://mathoverflow.net/a/414174/387190