Let $\Sigma$ be a connected noncompact orientable surface. I'm not assuming that $\Sigma$ is of finite type or anything -- for instance, I'm allowing $\Sigma$ to be the $2$-sphere minus a Cantor set. I'm pretty sure that the tangent bundle of $\Sigma$ is trivial. Here are two pieces of evidence.
This is true if $\Sigma$ can be embedded in a closed surface. Proof : it is easy to see that a closed surface minus a point has a trivial tangent bundle, so $\Sigma$ must have one too.
If $U\Sigma$ is the unit tangent bundle, then we have the standard short exact sequence $1 \longrightarrow \mathbb{Z} \longrightarrow \pi_1(U\Sigma) \longrightarrow \pi_1(\Sigma) \longrightarrow 1,$ where the $\mathbb{Z}$ is the loop around the fiber. Since $\Sigma$ is noncompact, the group $\pi_1(\Sigma)$ is free, so this short exact sequence splits (just like you would have if the tangent bundle was trivial).
Does anyone know how to prove this in general? Thanks!