Will's answer has the elements needed for a concrete reply to the question, but the question itself has caused some confusion about the setting and terminology which are worth clarifying.   First of all, the underlying field should be of characteristic different from 2, since it gets more subtle to talk about quadratic forms and orthogonal groups in characteristic 2.   (This is done however in work of Chevalley and Borel in the algebraic groups framework, where the groups are included in the classification.)

Originally the study of orthogonal groups was carried out (by Weyl, Chevalley, and many others) only in characteristic 0.  Here the (polynomial) condition on $n \times n$ matrices is just that the transpose of a matrix must equal its inverse.   The orthogonal matrices then form a group, say with the natural real or complex topology.   Orthogonal matrices have eigenvalues in these cases which are roots of unity paired with their inverses (or complex conjugates), allowing determinant $\pm 1$.   Thus the real or complex orthogonal group will have at least two connected components in the usual topology, considering the two cosets of the kernel of det.   The real or complex dimension of this Lie group is easily computed to be $\ell^2+\ell$ (in Lie theory terms, rank $\ell$ and $\ell^2$ positive roots).   It's worth following the case $n=5$ in Will's calculations: Lie type $B_2$. 

The special orthogonal group is the kernel of det here and is usually just called $\mathrm{SO}(n)$ or the like when the field is understood.  To show it is *connected* in the topological group setting, Chevalley in *Theory of Lie Groups* uses induction on $n$ and the characterization of the successive quotients as spheres.

Now in the Chevalley-Borel setting of linear algebraic groups (first over an algebraically closed field), not yet expressed in terms of group schemes, much of the previous study carries over with modifications.   For linear algebraic groups given the Zariski topology, irreducibility of the underlying variety fortunately coincides with connectedness in that coarse topology; the term  "connected" is then preferred.  The irreducible (= connected) components of an algebraic group are disjoint and equidimensional as well as finite in number (unlike some Lie groups): these are just the cosets of the identity component.  

Here the most general argument for connectedness of an algebraic group requires it to be generated by suitable irreducible subsets such as closed
connected subgroups.   In the framework of classical matrix groups, this is most easily done by generating the group by connected 1-dimensional unipotent groups.  With some care this approach even handles special orthogonal groups in characteristic 2.   

On the other hand, there is the possibility discussed in this question of appealing (in characteristic not 2) to a Cayley transform.   Here one is able to map isomorphically a nonempty open subset of an affine space (dense in the Zariski topology) onto a nonempty open subset of the matrix group in a concrete way.   Then it has to be seen, as Will shows, that in the specific situation of the special orthogonal group none of its hypothetical extra irreducible/connected components can lie in the excluded hypersurface given by nonvanishing of a determinant.  Dimension counting seems necessary here. 

The only source I can quote for this slightly esoteric approach is a terse exercise 2.2.2(2) in Springer's book *Linear Algebraic Groups*, where much is left to the reader's ingenuity.   (Are there earlier sources?)   Springer himself was attracted to this approach, I think, because he used the Cayley transform for classical groups to realize an isomorphism between unipotent and nilpotent varieties in the group and its Lie algebra.