= Solution
Let $\operatorname{Sym}_n$ be the vector space of real symmetric matrices and consider
$$
\Phi:M_n(\mathbb R)\longrightarrow\operatorname{Sym}_n,
\qquad
\Phi(A)=A^TA.
$$
At $R\in O(n)$,
$$
D_R\Phi(H)=R^TH+H^TR.
$$
This derivative is surjective: for $S\in\operatorname{Sym}_n$, choosing $H=RS/2$ gives $D_R\Phi(H)=S$. Thus $I$ is a regular value and the <orthogonal group> $O(n)=\Phi^{-1}(I)$ is a submanifold of codimension
$$
\dim\operatorname{Sym}_n=\frac{n(n+1)}2.
$$
The determinant takes only the values $\pm1$ on $O(n)$, so the <special orthogonal group> $SO(n)=\{R\in O(n):\det R=1\}$ is an open-and-closed submanifold of $O(n)$. Therefore
$$
\dim SO(n)=n^2-\frac{n(n+1)}2=\frac{n(n-1)}2.
$$
The tangent space is the kernel of $D_R\Phi$:
$$
\boxed{T_RSO(n)
=\{H:R^TH+H^TR=0\}
=\{RA:A\in\mathfrak{so}(n)\}},
$$
where the <Special orthogonal Lie algebra> $\mathfrak{so}(n)$ is the vector space of <skew-symmetric matrices>. This proves the <special orthogonal group as a submanifold> description.
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