I can see why this is true but I'm not sure how to prove it, any help would be appreciated.
Prove that the tangent space $TS^{n}_{x}$ at a point $x$ on the $n$-sphere $S^{n}:=\{x \in \mathbb{R}^{n+1} : \lVert x\rVert=1\}$ can be identified with the space $x^{\perp}=\{v \in \mathbb{R}^{n+1} : v \cdot x=0\}$.
Where the definition of tangent space is as follows: Let $M$ be a smooth $m$-manifold. Given $x \in M$, let $c_{0}, c_{1}$ be $C^{\infty}$ curves defined on open intervals $(a_{0},b_{0}), (a_{1},b_{1})$ with $0 \in (a_{0},b_{0})$ , $0 \in (a_{1},b_{1})$ and $c_{0}(0)=c_{1}(0)=x$. Define an equivalence relation $\sim$ by $c_{0} \sim c_{1}$ $\iff$ $\frac{d}{dt}(\psi_{U}\circ c_{0}(t))\mid_{t=0} = \frac{d}{dt}(\psi_{U}\circ c_{1}(t))\mid_{t=0}$ , where $\psi_{U}:U \rightarrow \psi_{U}(U)\subseteq \mathbb{R}^{m}$ is a chart. Define the tangent space $TM_{x}$ of $M$ at $x$ as the set of all equivalence classes $[c_{0}]$.