I am going through some notes on geometric mechanics. In the first section we let $A(t)$ denote some product of Euler rotations: $$ A(t)= \left( \begin{array}{ccc} \cos (\psi ) \cos (\varphi )-\cos (\theta ) \sin (\varphi ) \sin (\psi ) & \cos (\psi ) \sin (\varphi )+\cos (\theta ) \cos (\varphi ) \sin (\psi ) & \sin (\theta ) \sin (\psi ) \\ -\cos (\theta ) \cos (\psi ) \sin (\varphi )+\cos (\varphi ) (-\sin (\psi )) & \cos (\theta ) \cos (\varphi ) \cos (\psi )-\sin (\psi ) \sin (\varphi ) & \sin (\theta ) \cos (\psi ) \\ \sin (\theta ) \sin (\varphi ) & -\sin (\theta ) \cos (\varphi ) & \cos (\theta ) \end{array}\right), $$ where $0<\psi(t),\varphi(t)<2\pi$, $0<\theta(t)<\pi$ are Euler angles. This implies that $A(t)\in SO(3)$, the special orthogonal Lie group in three dimensions.
Later on we consider the quantities $$\omega_S(t)=\dot{A}(t)A^{-1}(t),$$ $$\omega_B(t)=A^{-1}(t)\dot{A}(t),$$ where the dot indicates the time derivative. We can verify through direct calculation that $\omega_S(t),\omega_B(t)\in\mathfrak{so}(3)$, the lie algebra of skew-symmetric matrices in three dimensions. Why does this happen?
Start with the defining property of the orthogonal group: that $A(t) A^T(t) = I$, where the $^T$ indicates the transpose. Suppressing the $t$, we get $AA^t = I$.
Take the derivative of this equation. You get $A' A^T + A (A^T)' = 0$.
Now, an easy calculation shows $(A^T)' = (A')^T$, so this turns into $A' A^T + A(A')^T = 0$. Using the fact that $(BC)^T = C^T B^T$, we can rewrite this as $A' A^T + (A' A^T)^T = 0$. In other words, $A'A^T$ is skew symmetric, so is an element of $\mathfrak{so}(n)$.
But, $A^T = A^{-1}$ for $A\in SO(n)$, so we have $A'A^{-1}\in \mathfrak{so}(n)$.