Tangent space of a sphere is locally homeomorphic to to $\mathbb{R}^{2n}$ 
Tangent space of $S^n$ is locally homeomorphic to to
$\mathbb{R}^{2n}$.

The tangent space is given by $$TS^n=\{(x,v)\in S^n\times R^{n+1} : v\perp x\}\subset S^n\times R^{n+1}$$
I didn't know where to start, so I tried researching. I found the "Whitney embedding theorem" that says:
The theorem states that every n-dimensional differentiable manifold has an embedding in $\mathbb{R}^{2n}$.
Is there a way of showing this without using this theorem?
 A: I think we can do this from scratch, without much differential geometry. For simplicity, consider $S^2.$ Since the sphere is rotationally invariant, without loss of generality, take $\vec p=(0,0,1).$ There is an open set $p\in U\subseteq S^2$ small enough so that if $\vec x\in U$ then the $z$ coordinate of $x$ is different from zero. i.e. $U$ does not intersect the equator.
Let $\phi:U\to \phi(U)$ be stereographic projection. $\ \phi $ is a homeomorphism.
If $\vec x=(x,y,z)\in U$ and $\vec v=(a,b,c)\in \mathbb R^3$ such that $\vec x \perp \vec v,\ $ then $ax+by+cz=0.$
Now, define the map
$\Phi:TU\to \phi(U)\times \mathbb R^2: (\vec x,\vec v)=((x,y,z),(a,b,c))\mapsto (\phi(\vec x), (a,b)).$
$\Phi$ is open, surjective and continuous so all we need to prove is that it is injective:
If $(\phi(\vec x_1),\vec v_1)=(\phi(\vec x_2),\vec v_2)$ then $\vec x_1=\vec x_2$ and $a_1=a_2$ and $b_1=b_2.$ But then
$a_1x_1+b_1y_1+c_1z_1=a_2x_2+b_2y_2+c_2z_2\Rightarrow$
$a_1x_1+b_1y_1+c_1z_1=a_1x_1+b_1y_1+c_2z_1\Rightarrow c_1=c_2.$
(In the last step we used the fact that $z_1\neq 0.$)
To finish, simply note that $\phi(U)\cong \mathbb R^2$ and $\mathbb R^4\cong \mathbb R^2\times \mathbb R^2.$
