I want to prove that $\xi$ is a rank $k$ vector bundle over an $n$ dimensional CW complex $X$ such that $k>n$. Then $\xi$ is trivial iff $\xi \oplus \varepsilon^1$ is trivial. Here $\varepsilon^1$ is the trivial bundle of rank $1$.
Note that one direction is trivial. The non trivial direction is if $\xi \oplus \varepsilon^1$ is trivial implies $\xi$ is trivial.
To prove this I want to use classifying spaces. Let $BO(k)$ be the classifying space of all rank $k$ vector bundles. Now let the bundle $\xi$ be given by the map $$f:X \to BO(k)$$ But we are given that $\xi \oplus \varepsilon^1$ is trivial. This is just pullback of the map $i \circ f$ where $i$ is the map $$i: BO(k) \to BO(k+1)$$ So now since $\xi \oplus \varepsilon^1$ is trivial $i \circ f$ is null homotopic. Now we want to show that $f$ is null homotopic. I am struck here. Any help will be appreciated. Thank you.
You can prove this using the long exact sequence associated to the pointed fiber sequence
$$ S^k = O(k+1)/O(k) \stackrel{j}{\to} BO(k) \stackrel{i}{\to} BO(k+1) $$
Namely, fix a basepoint $x_0$ in $X$ which is a $0$-cell. Then we have an exact sequence of pointed sets
$$ [X,S^k]_* \stackrel{j_*}{\to} [X,BO(k)]_* \stackrel{i_*}{\to} [X,BO(k+1)]_* $$
where $[Z,W]_*$ denotes the set of pointed homotopy classes of pointed maps from $Z$ to $W$, and the basepoint of $[Z,W]_*$ is the class of the constant map (the constant is of course the basepoint of $W$).
Using your notation, you have that $i \circ f$ is nullhomotopic. Since $BO(k+1)$ is path connected, we can assume that it is homotopic to the constant map $c$, where the constant is the basepoint of $BO(k+1)$. Moreover, since we chose $x_0$ to be a $0$-cell of $X$, we can assume that the homotopy preserves the basepoints (using the homotopy extension property).
This shows that $i_*[f] = [c]$ and $[c]$ is the basepoint of $[X,BO(k+1)]_*$, so there exists $g \colon X \to S^k$ pointed such that $[f]=j_*[g]=[jg]$. Since the dimension of $X$ is smaller than $k$, the map $g$ is nullhomotopic by cellular approximation, and by the same argument as before, it is pointed nullhomotopic.
Hence $f$ is pointed nullhomotopic, in particular nullhomotopic, and so $\xi$ is trivial.