= Solution
Consider $F(A)=A^TA$ with values in the symmetric <matrices>. At the identity,
$$
DF_I(X)=X^T+X,
$$
which is onto the $n(n+1)/2$-dimensional space of symmetric <matrices>. The same rank holds at every orthogonal <matrix>. Thus $O(n)=F^{-1}(I)$ is a manifold, and its determinant-one component, the <special orthogonal group>, has dimension
$$
n^2-\frac{n(n+1)}2=\frac{n(n-1)}2.
$$
The tangent equation at the identity is $X^T+X=0$, so
$$
\boxed{T_I SO(n)=\{X:X^T=-X\}.}
$$
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