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I recently learned that the only parallelizable spheres are $\mathbb{S}^1$, $\mathbb{S}^3$, and $\mathbb{S}^7$. This led me to wonder:
What is $T\mathbb{S}^2$? Is it diffeomorphic to a more familiar space? What about $T\mathbb{S}^n$ for $n \neq 1, 3, 7$?
EDIT (for precision): Is $T\mathbb{S}^2$ diffeomorphic to some finite product, connected sum, and/or quotient of spheres, projective spaces, euclidean spaces, and linear groups?
Your question is perhaps too vague. The tangent bundle has a definition and that's what it is -- presumably you're asking for more than this but you don't specify with any precision. It would be good to edit your question to make it more precise.
One description of $TS^2$ is you take the mapping cylinder $SO_3 \to S^2$ where this map is the orbit of a single point in $SO_3$'s natural action on $S^2$ by isometries. Technically, $TS^2$ is the above mapping cylindre after you erase the $SO_3$ boundary (to make it non-compact).
Another description of $TS^2$ would be the configuration space of two points in $S^2$. Precisely,
$C_2 S^2 = \{ (x,y) \in S^2 \times S^2 : x \neq y \}$
You can identify the two by a stereographic projection map construction. There's many more such constructions. But you really ought to say what you're looking for because the list is endless.
Here's how I think about it. (Ryan Budney posted his answer while I was typing this. One can think of this as a fleshing out of his answer).
First, we need to understand the unit tangent bundle $T^1 S^2$. Once we know this, we product this with $[0,\infty)$ and then quotient all points of the form $(u, 0)\in T^1S^2 \times\mathbb{R}$ somehow to get the 0 section $S^2$. (This is precisely the mapping cylinder construction Ryan mentions).
Before we can talk about the "unit tangent bundle", we must have a notion of length of vectors. So in the background, pretend like I picked a Riemannian metric so lengths make sense.
I claim $T^1 S^2$ is diffeomorphic to $SO(3)$ (the collection of 3 x 3 orthogonal matrices of determinant 1) which is diffeomorphic to $\mathbb{R}P^3$.
The map from $T^1 S^2$ to $SO(3)$ sends $(u,v)$ to the matrix with columns $u, v, u\times v$. Here, I'm thinking of a unit tangent vector $v\in T_u S^2$ as a vector in $\mathbb{R}^3$ orthogonal to the vector $u$.
The easiest way to see $SO(3)$ and $\mathbb{R}P^3$ are diffeomorphic is to note they are both quotients of $S^3 = SU(2)$ by the same quotienting map.
So, we understand $T^1 S^2$, the unit length vectors in $TS^2$.
To allow for length, we product with $[0,\infty)$. Now, the only problem is the 0 section should be an $S^2$, and it's currently an $\mathbb{R}P^3$, so some quotienting must happen.
What quotienting must happen? Well, all the unit vectors at a given point must collapse to the point. Well, there is the action of a circle on $T^1S^2$ given by rotation vectors clockwise (say) as seen from the normal vector to the sphere. This action is clearly free. Now, it's a fact that if you translate this circle action into the $SO(3)$ picture, the circle action is the Hopf action. This implies that we identify $\mathbb{R}P^3$ with $S^2$ by quotienting by the Hopf action: Two points in $\mathbb{R}P^3$ iff they are in the same Hopf orbit.
Incidentally, I just learned a few days ago that $TS^2$ is not homeomorphic to $S^2\times \mathbb{R}^2$, though I'm still not sure how to prove it ;-). (Of course, it's clear that they are not bundle isomorphic, but they could still be abstractly homeomorphic). I don't know about the other nonparallizable tangent bundles, though.
A slightly random answer, but if we concretely identify $TS^{n-1}$ as the embedded real manifold $\{ (x, v) \in \mathbb{R}^n \times \mathbb{R}^n : \| x \| = 1, \langle x, v \rangle = 0 \}$, then it is diffeomorphic to the complex affine quadric $\{ (z_1, \ldots, z_n) \in \mathbb{C}^n : z_1^2 + \cdots + z_n^2 = 1 \}$. (This was an amusing homework exercise I did yesterday.)
