It is well-known that if we have a "standard" smooth manifold (i.e., Hausdorff and second countable, and thus paracompact), $TM$ is isomorphic to $T^*M$. The usual argument follows from taking a Riemannian metric and using the musical isomorphism fibrewise.
What if the manifold is not paracompact? Is there an example of $M$ such that those bundles are not isomorphic? (The usual argument falls apart here.)
The long line is a "manifold" (but not second countable) which does not admit a Riemannian metric. There are smooth structures on this thing hence there is (given a smooth structure) a notion of tangent bundle and cotangent bundle. These are weird:
The long line is one dimensional, and an isomorphism $\phi:TL\rightarrow T^*L$ would therefore induce a Riemannian metric via the formula $g(X,Y)=\pm\phi(X)(Y)$ (You will need the minus sign if $\phi_p(X_p)(X_p)<0$ for a, hence all, non-zero $X_p$). Since no Riemannian metric exists, no such isomorphism can exist.