In the Lee‘s book there is a proposition stating:
If $M$ is a smooth $n$-manifold with or without boundary, and $M$ can be covered by a single smooth chart, then $TM$ is diffeomorphic to $M\times \mathbb{R}^n$.
This covers simple examples when $M$ is an open subset of $\mathbb R^n$. However, closed ball and closed annulus in $\mathbb R^n$ are not contained in the above statement.
My question is do we have (for any $n$)
- $TB_r \cong B_r\times \mathbb R^n$
- $TC(r_1,r_2) \cong C(r_1,r_2)\times \mathbb R^n$,
where $B_r:=\{x\in\mathbb R^n :|x|\leq r\}$ is closed ball, and $C(r_1,r_2):=\{x\in\mathbb R^n : r_1\leq|x|\leq r_2\}$ closed annulus.
By the stereographic projection, the problem with closed ball should be equivalent to one with closed hemisphere.
Look at the hypotheses: the closed ball IS a smooth $n$-manifold with boundary, covered by a single chart. The answer in each case you asked about is "yes". In fact, the map you get is not just a diffeomorphism, but a bundle map. Once you see this,
The annular case follows from the ball case, since if you have a map $$ F: TB(r_2) \to B_r^2 \times R^n $$ such that for $v \in TB(r_2)$, we have $$ \pi(v) = \pi_1 (F(v)) $$ where $\pi$ is the projection to the base space on the left, and $\pi_1$ is projection to the first factor on the right (i.e., it's a bundle map), you can simply restrict the domain of $F$ to $TC(r_1, r_2) \subset TB(r_2))$ to get a trivialization of $TC$.