Sub Tangent of a polar curve In the above diagram, the curve depicted in black is given by $$r = f(\theta)$$ and PT is the tangent and NP is the normal to the curve at P.

Suppose if $$P = (r,\theta)$$

How to prove below equation? $$tan(\psi) = r / (dr/d\theta)$$

Slope of a tangent in polar coordinate is given by $$\tan\gamma =m_{\text{tangent}}=\frac{r'\sin(\theta)+r\cos(\theta)}{r'\cos(\theta)-r\sin(\theta)}$$ Where $\gamma$ is the angle tangent makes with horizontal axis. $\gamma=\theta+\psi$ $$\tan\psi=\tan(\gamma-\theta)= \frac{\frac{r'\sin(\theta)+r\cos(\theta)}{r'\cos(\theta)-r\sin(\theta)}-\tan\theta}{1+\frac{r'\sin(\theta)+r\cos(\theta)}{r'\cos(\theta)-r\sin(\theta)}\tan\theta}=\frac{r}{r'}=\frac{r}{\frac{dr}{d\theta}}$$

Consider the blue differential ( representing infinitesmal) right angled triangle $PQR$ in a direct geometrical approach. Right angle is at $Q$.

Now $P$ has polar coordinates $(r, \theta),$ and

$$OP \approx OQ= r$$ Normal $PN$ can be considered parallel to $QR$ in the limit as $dr\rightarrow 0.$ Then,

$$\tan \psi= \dfrac{QR}{QP}=\dfrac{r \cdot d \theta}{dr} = \dfrac{r }{r^{'}} ;$$

Subtangent $OT= r \tan \psi = \dfrac{r^2}{r^{'}}$ and subnormal $ON= \dfrac{r}{ \tan \psi }= {r^{'}};$

Also if $QR$ makes $\angle \phi$ to x-axis, then external angle is sum of two internal angles:

$$\phi= \psi + \theta$$ 