I am trying to find the family of functions orthogonal to the family described by the differential equation, $$\frac{dr}{d \theta} = -\frac{r\sin{\theta}}{\cos{\theta}}$$ If this equation were expressed in rectangular coordinates, then I could swap $-\frac{dx}{dy}$ in for $\frac{dy}{dx}$ in the LHS to obtain the differential equation for the corresponding orthogonal family. Unfortunately, since $\frac{dr}{d \theta}$ does not correspond to the slope of the tangent line of the curve at $(r, \theta)$, this can't be solved this way.
Instead, let $\psi = \gamma - \theta$ denote the angle between the tangent, $\gamma$, and the radius, $\theta$. We have that $\tan{\psi} = r \frac{d \theta}{dr}$. I am trying to use this fact to relate $r$ and $\frac{dr}{d \theta}$ with the slope of the tangent. Since the slope of the tangent line to the curve is $\tan{\gamma}$ we can write, \begin{align*} \tan{\gamma} &= \tan{\psi + \theta} = \frac{\tan{\psi} + \tan{\theta}}{1 - \tan{\psi}\tan{\theta}} \\ \end{align*}
but I'm not sure where to go from here.
The book I am using says that since $\tan{\psi} = r\frac{d \theta}{dr}$, we can replace $r\frac{d \theta}{dr}$ with its negative reciprocal $-\frac{1}{r}\frac{dr}{d \theta}$ to get, $$r \frac{d \theta}{dr} = \frac{\sin \theta}{\cos \theta}$$ However, this seems wrong to me since the slope of the tangent line at $(r, \theta)$ is $\tan{\gamma}$, not $\tan{\psi}$; why does it seem like the book is suggesting that the slope is $\tan{\psi}$? What am I missing here?
Thanks!