I know that the tangent bundle $TS^2$ is stably trivial: if $\nu$ denotes the normal bundle of the embedding of $S^2$ in $\mathbb{R}^3$ as the unit sphere, then $\nu$ is a trivial line bundle. But then the sum of these is trivial, since it just gives the restriction of the tangent bundle on $\mathbb{R}^3$. If I wanted, I could just tack on an extra trivial line bundle to have $TS^2 \oplus \varepsilon^2 \cong \varepsilon^4$.
Now $S^2 \cong \mathbb{CP}^1$ is also a complex manifold, so $TS^2$ is a complex line bundle. If this complex bundle were stably trivial, then the product theorem for Chern classes would yield that the total Chern class of $TS^2$ is trivial. However, $$c(TS^2) = 1 + c_1(TS^2) = 1+e(TS^2)$$ where the Euler class $e(TS^2)$ is twice a generator of $H^2(S^2) \cong \mathbb{Z}$, a contradiction. Doesn't this now contradict the last line of the previous paragraph, since rank-$2k$ real trivial bundles are the same as rank $k$-complex trivial bundles?