How do I show that the rank of the following matrix is 2?
$\begin{pmatrix} -(a+b\cos(\phi))\sin(\theta) & -b\sin(\phi)\cos(\theta) \\ (a+b\cos(\phi))\cos(\theta) & -b\sin(\phi)\sin(\theta) \\ 0 & b\cos(\phi) \end{pmatrix}$
Where $\theta, \phi \in \mathbb{R}$ and $0<b<a$
Let's call your matrix $A$, with elements $A_{i,j}$. $A$ has dimension $3\times2$, hence the rank is at most $2$. We will prove it's effectively $2$ by exhibiting two linearly independant rows among the three, or equivalently a regular $2\times2$ submatrix.
The determinant of the first two rows is $b\sin\phi(a+b\cos \phi)(\cos^2 \theta+\sin^2\theta)=b\sin\phi(a+b\cos \phi)$.
Considering your constraints, it's nonzero (thus $A$ has rank $2$) unless $\phi=k\pi$. Notice $a+b\cos\phi$ can't be zero since $a>b\geq |b\cos\phi|$.
And when $\phi=k\pi$, $A_{1,2}=A_{2,2}=0$ and one of $A_{1,1}$ or $A_{2,1}$ is certainly nonzero (sine and cosine do not vanish for the same $\theta$, and the factor $a+b\cos\phi$ is nonzero). Since $A_{3,2}\neq0$ as well, you have then a $2\times2$ submatrix which is diagonal (either rows $2$ and $3$ of $A$, either rows 1 and 3), with nonzero diagonal elements, hence rank $2$.