In general, there exist non-trivial orientable bundles. But if we only consider line bundles, then orientable bundles are always trivial bundles. (Similarly, non-trivial bundles are always non-orientable bundles.) Is this statement true?
I got this conclusion from the following question and answer: If a line bundle admits a non-vanishing section then it is trivial. If a line bundle admits a non-vanishing section, then it conforms to the definitions of trivial bundle and orientable bundle at the same time.
Let $(E,B,L)$ be an orientable line bundle, this means there exists trivializations $U_{i}$ such that $\forall i; \phi_{U_{i}}:\pi^{-1}(U_{i}) \mapsto U\times \mathbb{R}$ is fiberwise orientation preserving, denote by $g_{i}$ the inverse of this map, pick a partition of unity $P_{i}$ of the base space subbordinated to the trivializations (such a partition exists whenever $B$ is assumed to be paracompact). Define the map $s:B\mapsto E$ by the formula $s(b)=\sum P_{i}(b) g_{i}(b,1)$ where the sum is (necessarily finite) and is over the $U_{i}$ to which $b$ belongs.
Remark that $\forall b\in B,s(b)$ is never zero because preserving orientation between a one dimensional vector space and $\mathbb{R}$ means that $g_{i}(b,1)$ is always one "the same side" for every $b$ and every $I$.
This proves that an orientable line bundle is necessarly trivial.the other direction is easy.