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?
