Prove that $TS^{n-1}$ is a trivial bundle, if $\mathbb{R}^n$ may be provided with an $\mathbb{R}$-algebra structure without zero divisors Here I was shown how to prove that $TS^1$ is a trivial bundle. 
Similarly, I can show that $TS^3$ is a trivial bundle.
Identify $\mathbb{H}$ with $\mathbb{R}^4$ and that that for $x \in S^3$ we have $x \mapsto ix$, $x \mapsto jx$ and $x \mapsto kx$ as orthogonal vectors, which are also orthogonal to $x$. Then we can see that there is an ismorphism $S^3 \times \mathbb{R}^4 \to S^3 \times \mathbb{R}^3$ given by $(x,z) \mapsto (x,iz_1/x \times iz_2/x \times iz_3/x)$ where $z_1 = \{ t_1 i x| t_1 \in  \mathbb{R} \},z_2 = \{ t_2 j x| t_2 \in  \mathbb{R} \},z_3 = \{ t_3 k x| t_3 \in  \mathbb{R} \}$ (Possibly I haven't written that last bit so nicely).
It is also possibly to do similar construction for $S^7$ using Octonions. 
This then leads onto the general question:

Prove that $TS^{n-1}$ is a trivial bundle, if $\mathbb{R}^n$ may be provided with an $\mathbb{R}$-algebra structure without zero divisors.

Of course, now my construction kind of fails - I can't very well construct the map $x \mapsto ix$ (for example), because I don't have a multiplication table! So what is the more general way to approach this problem? (From a simple algebra approach - let's not invoke Bott periodicity or anything just yet!)
Edit: Maybe the question should be - What property of $\mathbb{R}$-algebra structures without zero divisors, helps to solve this problem?
Edit 2: (The following is probably wrong). So assume that $\mathbb{R}^n$ has an $\mathbb{R}$ algebraic structure without zero divisors. We seek an isomorphism $$\phi:S^{n-1} \times \mathbb{R}^n \to S^{n-1} \times \mathbb{R}^{n-1}.$$ Denote our algebraic structure by $\mathbb{T}$, and let it has orthonormal basis $\{ e_0, \cdots, e_{n-1} \}$. Then $S^{n-1}$ is the subset of $\mathbb{T}$ with (Euclidean norm) = 1 (I think I can still define the Euclidean norm?). Then given $z \in S^{n-1}$, $\{ ze_0, \cdots, ze_{n-1} \}$ is still an orthonormal basis (maybe). Then define the function 
$$(x,z) \mapsto (x,iz_1/x \times \cdots\times iz_{n-1}/x)$$ where $z_i = \{ t_i e_i x| t_i \in  \mathbb{R} \}$
(the division exists because $x$ is orthogonal to $z$ and we have no zero divisors)
 A: Suppose that $\mathbb R^n$, with basis $(e_1,...,e_n)$, has the structure of a division ring  and denote by $a.b$ the product of the vectors $a$ and $b$. Consider  $x\in S^{n-1}$; then the vectors $e_1.x, ...,e_n.x$ are linearly independant (cf. right multiplication by $x^{-1}$) and you can apply the Gram-Schmidt process to them, obtaining $v_1(x),...,v_n(x)$ with $v_1(x)=e_1.x/||e_1.x||$.  The vectors $v_2(x),...,v_n(x)$ are an orthonormal basis of the tangent space to $S^{n-1}$  at the point $v_1(x)$ and so the $(n-1)$ vectors $ v_i(v_1^{-1}(x) \quad (i=2,...,n)$ are an orthonormal basis of the tangent space $T_x(S^{n-1})$ of the sphere at $x$. This is the desired trivialization of the tangent bundle $TS^{n-1} $.
Remarks
1) To be perfectly explicit , the formula for $v_1^{-1}$ is 
$ v_1^{-1}(y)=e_1^{-1}.y/||e_1^{-1}.y||$.
2) The euclidean scalar product and norm are the usual ones on $\mathbb R^n$ and have absolutely no interference with the division algebra structure: let us be sure not to confuse the vector $a.b \in \mathbb R^n$ and the scalar product $< a, b>\in \mathbb R$  !
3) However the quaternions have a supplementary structure: the conjugation which is an $\mathbb R$-linear anti-involution. This allows the quaternions to see the the euclidean structure of $\mathbb R^4$ thanks to the pleasant formula
linking the division ring structure and the metric structure
 $$v.\bar v=||v||^2 1_{\mathbb H} \quad ( v\in \mathbb H)$$
