Understanding why the tangent bundle of a unit circle is diffeomorphic to $S^1\times\mathbb R$ Wikipedia  defines a tangent bundle of a manifold $M$ as
$$\coprod_{x\in M}T_{x}M = \{(x,y)| x\in M, y\in T_x M\} .$$ 
The wiki link describes the tangent bundle of $S_1$ as being diffeomorphic to $S_1\times R$ by saying the way to make the tangent spaces at all points of the circle (which are in the plane of the circle) disjoint smoothly is by aligning them perpendicular to the plane of the circle which I don't understand completely. 
The tangent space at a point on the circle, is just the tangent line at that point and the line is not tangent to any other point on the circle, so aren't the tangent spaces disjoint to begin with? Thanks and appreciate a hint.
 A: The tangent space to a manifold $M$ is a disjoint union like you say: More precisely, if a vector $X \in TM$ is in the tangent space $T_p M$ to $p$, it's not in $T_q M$ for any $q \neq p$.
Now, if a manifold $M$ is viewed as a submanifold of $\Bbb R^n$, it is sometimes useful to identify the tangent space $T_p M$ as an affine subspace of $\Bbb R^n$ tangent to $M$ at $p$. For example, we might identify the tangent space $T_z S^1$ at $z \in S^1 \subset \Bbb R^2$ with $\{w : z \cdot (w - z) = 0\}$. This identification, however, can lead to confusion, and this confusion is exactly what the comment in the Wikipedia article aims to prevent: For $z' \in S^1$, $z' \neq \pm z$, the affine spaces in $\Bbb R^2$ we identify with $T_z S^1$ and $T_{z'} S^1$ intersect, even though $T_z S^1$ and $T_{z'} S^1$ are themselves disjoint.
If we view $\Bbb R^2$ as a plane in $\Bbb R^3$, one can imagine rotating each such affine line in $\Bbb R^2$ a quarter turn to be orthogonal to the plane of the circle; then, these lines are pairwise disjoint (hence resolving the above issue) and comprise a cylinder ($S^1 \times \Bbb R$) that we can identify with the tangent space $T S^1$. (This is perhaps misleading in its own way, but if it helps you, great.)
We can make the identification $TS^1 \leftrightarrow S^1 \times \Bbb R$ in another way, by the way: If we identify $\Bbb R^2 \cong \Bbb C$, then as above we have a canonical identifiaction $T_1 \Bbb C \cong \Bbb C$. Then identifying $$S^1 = \{z \in \Bbb C : |z| = 1\}$$ identifies $T_1 S^1$ with the imaginary axis $i \Bbb R \subset \Bbb C$. By symmetry, for any $z \in S^1$, the rotation of $S^1$ that takes $z$ to $1$ (namely, multiplication by $z^{-1} = \bar z$) maps $T_{z'} S^1$, again viewed as an affine space in $\Bbb C$ to $T_1 S^1$. This in fact defines a diffeomorphism $TS^1 \cong S^1 \times i \Bbb R$ by
$(z, X) \mapsto (z, z^{-1} X)$, where we regard $X$ as an element of $T_z \Bbb C \cong \Bbb C$. Of course, $i \Bbb R$ and $\Bbb R$ are canonically isomorphic given by the map that divides by $i$, defining an isomorphism
$$TS^1 \cong S^1 \times \Bbb R, \qquad (z, X) \mapsto (z, -i z^{-1} X) .$$
Remark Here we took advantage of the fact that $S^1$ can be viewed as a Lie group and let it act on itself on the left. Essentially the same construction gives for any Lie group $G$ an explicit diffeomorphism $TG \cong G \times \operatorname{Lie}(G)$, which in particular shows that every Lie group is parallelizable.
