2 added 7 characters in body

Each connected component of an algebraic group has the same dimension. Thus if it has a connected subvariety whose complement has a lower dimension, it is connected. (If it had two connected components, only one could contain the connected subvariety, and so it the other would have a lower dimension.)

The subvariety where $I_n+A$ is invertible is birational to $\mathbb A^{\frac{n(n-1)}{2}}$ and thus has dimension $\frac{n(n-1)}{2}$.

The complement is where $A$ has a $-1$ eigenvalue. But since every eigenvalue to an orthogonal matrix must have its inverse also an eigenvalue, the determinant is the product of all the eigenvalues which are their own inverse, which are just $-1$ and $1$. Thus, if the determinant is $1$, the number of $-1$ eigenvalues is even, so is at least $2$. We can split the matrix into a $2$-dimensional $-1$-eigenspace and a matrix in $SO(n-2)$. These things are paramaterized by a $SO_{n-2}$-bundle on $G_{2}^{n-2}$, whose dimension is $2(n-2)+\frac{(n-2)(n-3)}{2}=\frac{n(n-1)}{2}-1$. If the $-1$-eigenspace is more than two-dimensional there is more than one way to express a matrix in this way, but that can only decrease the dimension.

1

Each connected component of an algebraic group has the same dimension. Thus if it has a connected subvariety whose complement has a lower dimension, it is connected. (If it had two connected components, only one could contain the connected subvariety, and so it would have a lower dimension.)

The subvariety where $I_n+A$ is invertible is birational to $\mathbb A^{\frac{n(n-1)}{2}}$ and thus has dimension $\frac{n(n-1)}{2}$.

The complement is where $A$ has a $-1$ eigenvalue. But since every eigenvalue to an orthogonal matrix must have its inverse also an eigenvalue, the determinant is the product of all the eigenvalues which are their own inverse, which are just $-1$ and $1$. Thus, if the determinant is $1$, the number of $-1$ eigenvalues is even, so is at least $2$. We can split the matrix into a $2$-dimensional $-1$-eigenspace and a matrix in $SO(n-2)$. These things are paramaterized by a $SO_{n-2}$-bundle on $G_{2}^{n-2}$, whose dimension is $2(n-2)+\frac{(n-2)(n-3)}{2}=\frac{n(n-1)}{2}-1$. If the $-1$-eigenspace is more than two-dimensional there is more than one way to express a matrix in this way, but that can only decrease the dimension.