The tangent space $T_x \mathbb S^n$ consists of such points in $\mathbb R^{n+1}$ that are perpendicular to $x$

Solution 1:

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.