By homotopy invariance of the de Rham cohomology and the fact that the total space of every smooth vector bundle $E\to M$ is homotopy equivalent to its base space $M$ ($M$ is a smooth manifold), we have the following isomorphism of singular cohomologies:
$$H^p(E, \mathbb C)\cong H^p(M, \mathbb C)$$
for every $p=0,.., \max\{\dim E, \dim M\}$. Does this mean that the top singular cohomology degree of the tangent bundle $TM$ is $\dim M$? Meaning,
$$H^p(TM, \mathbb C)=0$$ for all $p>\dim M$?