Proving that $TS^n \oplus (S^n \times \mathbb{R}) \cong S^n \times \mathbb{R}^{n+1}$ as vector bundles. I was going through some exercise to prepare for an upcoming exam and I stumbled upon the following one, I am not fully able to solve.

Exercise 6. Consider the tangent bundle $TS^n$ of the $n$-dimensional sphere. Show that the direct sum of $TS^n$ with the trivial bundle of rank one is isomorphic to the trivial vector bundle of rank $n+1$. Note: This does not imply that $TS^n$ is isomorphic to a trivial bundle.

My Own Attempts
Little remark before I begin:

In the manual I am using, we call two vector bundles $E$ and $F$ isomorphic if there exists a collection of maps $\{u_x \colon E_x \to F_x\}_{x \in M}$　that are isomorphisms as vector spaces such that the induced map $U \colon E \to F$ is smooth.

Using this definition I proved that a vector bundle is trivial if and only if you can find a global frame. So my goal was to find a global frame of $TS^n \oplus (S^n \oplus \mathbb{R})$.
I had a bit of trouble doing this, so I decided to look at the fibres:
$$
(TS^n \oplus (S^n \oplus \mathbb{R}))_x \cong T_x S^n \oplus \mathbb{R}, \quad (S^n \oplus \mathbb{R}^{n+1})_{x} \cong \mathbb{R}^{n+1}.
$$
Without proving it, I sort of intuitively thought that
$$
T_x S^n = \{v \in \mathbb{R}^{n+1} \mid \langle v, x \rangle = 0\},
$$
where $\langle \cdot, \cdot \rangle$ denotes the inner product on the Euclidean space. I wasn't directly able to prove it, but I found another stackexchange post confirming this train of thought. I hoped I was able to then to be able to construct a global section using this description, but I was unfortunately not able to do so.
Other stackexchange posts
After these attempts, I tried to find out whether I could gather some hints by looking at some other stackexchange posts. First I came across this post, in which it is stated "the sum of the tangent bundle and normal bundle of the sphere is trivial." I have personally never worked with normal bundles, but I think that the proof I have to deliver is a simpler version of this proof. I went on looking, but I found other posts that seemed to concern that exact same point (e.g. this one, and this one).
Concluding
I hope that I have provided enough resources for you to identify where I stand with this problem. If it isn't too much trouble, I would like some tips on how to further tackle this exercise.
Thanks a lot in advance!
P.S. If I need to specify more definitions, don't be afraid to ask!
 A: Thanks to Tom Ariel's sharp insight, I was set on the right path again.
The question is flawed, we should actually want to prove that
$$
TS^n \oplus \mathbb{R} \cong S^{n} \oplus \mathbb{R}^{n+1}.
$$
See the comments under the original question to see why I made this initial mistake.
The strategy I used now is to explicitly write out what both sides look like.
We can view
$$
TS^n = \{(x,v) \mid \langle v, x \rangle = 0, x \in S^n,  v \in \mathbb{R}^{n+1} \},
$$
Now using that any vector $w \in R^{n+1}$ can be uniquely written as a sum of vectors tangent to and normal the sphere (see comments in this post), we explicitly write down the map
$$
S^n \times \mathbb{R}^{n+1} \to TS^n \times \mathbb{R}: (x,v) = (x, v - \langle v,x \rangle x, \langle x, v \rangle).
$$
You can easily check that this map is well-defined:
$$
\langle x, v - \langle v,x \rangle x \rangle = \langle x, v \rangle - \langle v,x \rangle \langle x, x \rangle = 0,
$$
where in the last step we used that $x \in S^n$. It is (I think) relatively easy to show that this map is bijective by using the fact that any vector $w \in R^{n+1}$ can be uniquely written as a sum of vectors tangent to and normal the sphere. I also think we can claim that this map is smooth as the inner-product is a smooth function and the total function is thus a composition of smooth maps. It is also easy to see that this map is compatible with the fibers of both vector spaces (as it preserves the base point $x$). Using this, we can safely assume that the two spaces are isomorphic as vector bundles.
