Tangent Bundle is trivial implies that Cotangent Bundle is trivial.

Question

Let $M^{n}$ n-dimensional smooth manifold. Show that if $TM$ is trivial, then $T^{*}M$ is trivial. My idea: From a smooth global frame for $T^{*}M$ we say, $(E_{1},...,E_{n})$, in each $p\in M$, $(E_{1}(p),...,E_{n}(p))$ is basis for $T^{*}_{p}M$, Which we can consider the associated dual basis $(e_{1}(p),...,e_{n}(p))$. Making identification of $(T_{p}M)^{**}$ with $T_{p}M$(since that $dim(T_{p}M)=n$ is finite), we can consider $(e_{1}(p),...,e_{n}(p))$ basis for $T_{p}M$ with dual basis $(E_{1}(p),...,E_{n}(p))$. Define $(e_{1},...,e_{n})$ as above, os true that $e_{i}, i=1,...,n$ are smooth sections(in the case, smooth vector fields)?