In our lectrue we consider the function $f:\{(r,\varphi,\theta)\in\mathbb{R}^3\mid r>0\} \to\mathbb{R}^3$, where \begin{align*} &f(r,\varphi,\theta):=\begin{pmatrix}r\cos(\varphi)\cos(\theta)\\r\sin(\varphi)\cos(\theta)\\r\sin(\theta)\end{pmatrix}\\ &Df=\begin{pmatrix}\cos(\varphi)\cos(\theta)&-r \sin(\varphi)\cos(\theta)&-r\cos(\varphi)\sin(\theta)\\\sin(\varphi)\cos(\theta)&r\cos(\varphi)\cos(\theta)&-r\sin(\varphi)\sin(\theta)\\\sin(\theta)&0&r\cos(\theta)\end{pmatrix} \end{align*} We must check if rank$(Df)=3$ by computing the reduced row echelon form. Our solution says
"If $\cos(\theta)=0$ then $Df=\begin{pmatrix}1&0&0\\0&0&1\\0&0&0\end{pmatrix}$ and If $\cos(\theta)\neq0$ then $Df=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$".
Only if I assume $\cos(\theta)\neq0$ and $\sin(\varphi)\neq 0$, $\sin(\theta)\neq 0$, $\cos(\varphi)\neq0$, then after some tedious calculation I get $$Df=\begin{pmatrix}1&0&0\\0&\sin(\varphi)\cos(\varphi)\cos^2(\theta)&0\\0&0&\sin(\theta)\cos(\theta)\end{pmatrix}\implies Df=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$
Only if I assume $\cos(\theta)=0$ and $\sin(\varphi)\neq 0$, $\cos(\varphi)\neq0$ I get $$ Df=\begin{pmatrix}1&0&0\\0&0&1\\0&0&0\end{pmatrix}. $$
I am not sure if further assumptions regarding $\sin(\varphi)$, $\sin(\theta)$ and $\cos(\varphi)$ are missing in the solution or if there is a trick to get rid of $\sin(\theta)$, $\sin(\varphi)$, $\sin(\theta)$ and $\cos(\varphi)$ without further assumptions?