Going From a Matrix to Eulers When Cosine is 0

Question

Given a rotation matrix created from the Eulers $\phi$, $\theta$, $\psi$ like this which I want to take back to Eulers: $$ \begin{bmatrix} \cos\theta\cos\psi & -\cos\theta\sin\psi & \sin\theta \\ \sin\phi\sin\theta\cos\psi + \cos\phi\sin\psi & -\sin\phi\sin\theta\sin\psi + \cos\phi\cos\psi & -\sin\phi\cos\theta \\ -\cos\phi\sin\theta\cos\psi + \sin\phi\sin\psi & \cos\phi\sin\theta\sin\psi + \sin\phi\cos\psi & \cos\phi\cos\theta \end{bmatrix} $$ I can easily get $\theta$ from the top right component. But what happens when the top right is 1 or negative 1? That means that the rest of the top row and top column will go to 0. In the instance of 1, taking my matrix to: $$ \begin{bmatrix} 0 & 0 & 1 \\ \sin\phi\cos\psi + \cos\phi\sin\psi & -\sin\phi\sin\psi + \cos\phi\cos\psi & 0 \\ -\cos\phi\cos\psi + \sin\phi\sin\psi & \cos\phi\sin\psi + \sin\phi\cos\psi & 0 \end{bmatrix} $$ I know that I can use the sum and difference formulas to reduce this to: $$ \begin{bmatrix} 0 & 0 & 1 \\ \sin(\phi +\psi) & \cos(\phi + \psi) & 0 \\ -\cos(\phi + \psi) & \sin(\phi + \psi) & 0 \end{bmatrix} $$ I've read here in section 2.1 that this is indeterminate but that: $$ \phi + \psi = \tan^{-1}(\sin(\phi +\psi) / \cos(\phi + \psi)) $$ I'm having trouble making that step. How did they go from that last matrix to the equality?

Answer 1

The last step follows from the definition of the tangent :

$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$