I'm trying to prove the characteristic of the tangent space of a $n$-sphere. Could you have a check on my proof?
Theorem: Consider the $n$-sphere $\mathbb S^n \subseteq \mathbb R^{n+1}$ and $x \in \mathbb S^n$. Then $T_x \mathbb S^n$ consists of such points in $\mathbb R^{n+1}$ that are perpendicular to $x$.
Proof: First, $T_x \mathbb S^n$ is a $n$-dimensional vector subspace of $\mathbb R^{n+1}$. Wlog, we assume $x_{n+1}>0$. Let $S:= \{x \in \mathbb S^n \mid x_{n+1}>0\}$. Then $S$ is open in $\mathbb S^n$ and $x \in S$. Consider the map $$\varphi: \mathbb B_{\mathbb R^{n}} (0,1) \to S, y \mapsto \left (y_1, \ldots, y_n, \sqrt{1-(y_1^2+\cdots+ y_n^2)} \right ).$$
Then $\varphi$ is a local parameterization around $x$. Let $y := \varphi^{-1} (x)$ and $\varphi^i$ be the $i$-th coordinate of $\varphi$. Then $\mathrm d \varphi_y^i: \mathbb R^{n} \to \mathbb R, z \mapsto z_i$ for all $i=1, \ldots, n$. The partial derivative of $\varphi^{n+1}$ w.r.t. the $i$-th coordinate of $y$ is $$\frac{\partial \varphi^{n+1}}{\partial y_i}= - \frac{y_i}{\sqrt{1-(y_1^2+\cdots+ y_n^2)}}, \quad i=1,\ldots,n.$$
Then $$\mathrm d \varphi^{n+1}_{y}: \mathbb R^{n} \to \mathbb R, z \mapsto - \frac{z_1y_1 + \cdots+z_ny_n}{\sqrt{1-(y_1^2+\cdots+ y_n^2)}}.$$
Hence $$\mathrm d \varphi_y: \mathbb R^{n} \to \mathbb R^{n+1}, z \mapsto \left (z_1, \ldots, z_n, - \frac{z_1y_1 + \cdots+z_ny_n}{\sqrt{1-(y_1^2+\cdots+ y_n^2)}} \right ).$$
It follows from $y = \varphi^{-1} (x)$ that $x_{n+1} = \sqrt{1-(y_1^2+\cdots+ y_n^2)}$ and $x_i = y_i$ for all $i=1,\ldots,n$. Thus $$\langle \mathrm d \varphi_y (z), x \rangle = z_1x_1 + \cdots + z_nx_n - \frac{z_1y_1 + \cdots+z_ny_n}{\sqrt{1-(y_1^2+\cdots+ y_n^2)}}x_{n+1} = 0.$$
This completes the proof.
It seems correct to me. Note that you also have a coordinate-free proof using the fact that $$\mathbb{S}^n=\{x\in\mathbb{R}^{n+1}\,;\,||x||^2=1\},$$ thus for any smooth curve $\gamma:\mathbb{R}\to \mathbb{S}^n$ with $\gamma(0)=p$ and $\gamma'(0)=v$ you will have that $$\left.\frac{\partial}{\partial t}\right|_{t=0}||\gamma(t)||^2=2\langle\gamma(0),\gamma'(0)\rangle=2\langle p,v\rangle=0,$$ which shows that $T_p\mathbb{S}^n\subset\{v\in\mathbb{R}^{n+1}\,;\,\langle p,v\rangle=0\}$. Since the dimensions of these two vector spaces agree, the last inclusion is actually an equality.