If A and B are real orthogonal matrices how to prove that either A-B or A+B is singular?

Question

Degree of matrices is odd $n$-th degree. I figured out all eigenvalues of matrices A and B have to $1$ or $-1$. Now I assume I have to prove $\det((A-B)(A+B)) = 0$ and from that either $\det(A-B)$ or $\det(A+B)$ has to be $0$. How can I do that?

Answer 1

Hint.

$\det(A+B) \det(A-B) =\det(A+B) \det(A^T-B^T) = \det((A+B)(A^T-B^T)) $

Also note that for an odd-dimensional antisymetric matrix $M$, $\det(M)=0$.