Finding a basis of the tangent Space of the unit sphere Before I begin, there has been a similar question asked here before: Basis of tangent space of a sphere. However, I wanted to see if my attempt was correct (I use an alternative method not used by any of the other answers).
Let $S^2$ be the unit sphere (that is the one in $\mathbb{R}^3$) of dimension 2. I am attempting to find a basis of $S^2$'s tangent space.
Let $p$ be an arbitrary point in $S^2$. Then, we can represent $p$ via $(\sin{\theta}\cdot\cos{\varphi},\sin{\theta}\cdot\sin{\varphi},\cos{\theta})$ where $0\leq\theta<\pi$ and $0<\varphi\leq2\pi$. Thus, we have a mapping 
$$\phi(\theta,\varphi)=(\sin{\theta}\cdot\cos{\varphi},\sin{\theta}\cdot\sin{\varphi},\cos{\theta})$$
Then, note that:
$$\phi^{'}(\theta,\varphi) = \begin{pmatrix} \cos{\theta}\cdot\cos{\varphi} & \cos{\theta}\cdot\sin{\varphi} & -\sin{\theta} \\ -\sin{\theta}\cdot\sin{\varphi} & \sin{\theta}\cdot\cos{\varphi} & 0\end{pmatrix}$$
My question is that, does this resultant matrix span the tangent space of $S^2$? I am fairly sure that if it is true, then because there are two spanning vectors and $S^2$ is two-dimensional, it is a basis of the tangent space of $S^2$.
 A: Yes this is true, except for $\theta \neq 2\pi, \pi$ as mentioned in comments, which corresponds to North and South Pole, because your matrix is zero.  For see why this is true for the other points, you can simply say fix $\phi$ to be constant and vary $\theta$. You will get a path $\gamma(\theta)$, and its derivative is by definition an element of $T_pS^2$ but this is precisely the first row of your matrix. Changing the role, you can see that the second row is also in the tangent space by fixing $\theta$ this time.
Again, as mentionned in comment, if your solution was working for any angles this will gives a frame for $TS^2$, i.e two non-vanishing vector fields on the sphere. It is well known that such vector field does not exists, e.g using the Poincaré-Hopf theorem. This does not mean that there is no tangent space at the North pole or South pole, but just that you need another parametrization. 
Notice that the space of all the tangent vectors to $S^2$ have a really nice description : $TS^2 = \{(p,v) \in S^2 \times \Bbb R^3 : \langle p, v \rangle = 0 \}$.
