Let $G:=GL^+(n)$ (all invertible $n \times n$-matrices with positive determinant) and $K:=SO(n)$. Let $\mathfrak{g}, \mathfrak{k}$ denote their Lie algebras and $E:=\mathfrak{g}/\mathfrak{k}$. I would like to understand
Why the (total space of) tangent space $T(G/K)$ of $G/K$ can be identified with $(G \times E)/K$.
Some facts which are familiar for me:
0. $K$ acts on $E$ via adjoint representation and on $G$ via group multiplication. Thus one has a diagonal action of $K$ on $G \times E$ i.e. $a \cdot (g,x):=(ag,ad(a)x)$. I guess that $(G \times E)/K$ should be understood using this action.
1. if $G$ acts in such a way that $M/G$ is a manifold then we have a canonical identification for $x \in M$:
$$T_xM/T_x(G \cdot x) \cong T_{\pi(x)}(M/G)$$ where $G \cdot x$ is the orbit of $x$ and $\pi$ is the canonical projection.
In our case we have for $g \in G$:
$$T_gG/T_g(K \cdot g) \cong T_{\pi(g)}(G/K).$$
Since this correspondence is canonical it should give rise to vector bundles isomorphism (but which bundles?). 2. Lie groups are parallelisable thus $TG \cong G \times \mathfrak{g}, TK \cong K \times \mathfrak{k}$.
HINT: Define $\Phi\colon G\times E \to T(G/H)$ by $\Phi(g,v) = \big(\pi(g),\pi_*(L_{g*}v)\big)$. Check that this induces a well-defined isomorphism on the quotient.
EDIT: In fact, to get well-definedness, it's important to get the action exactly right. Note that $K$ should act on $G$ on the right (so that you get a well-defined coset), and so I believe the correct action should be $a\cdot(g,x) = (ga,\text{ad}(a^{-1})x)$. Then note that $$\Phi(ga,\text{ad}(a^{-1})v) = \big(\pi(ga),\pi_*(L_{(ga)*}\text{ad}(a^{-1})v)\big) = \big(\pi(g),\pi_*(L_{g*}L_{a*}L_{a^{-1}*}R_{a*}v)\big) = \big(\pi(g),\pi_*(L_{g*}v)\big),$$ as desired.