So in the picture above it states that $\tan(\theta) = \dfrac{r}{\tfrac{dr}{d \theta}}$and I’m asked to show that this is true. So what I did was I used the identity $\tan(x) = \sin(x)/\cos(x)$ and so that means $\sin(\theta)= r$ and $r'=\cos(\theta)$. Am I correct cause I have no clue if what I’m doing is even relevant.
Thanks in advanced :)
What we want to do is to make everything as polar coordinates, e.g. $r, \theta, \frac{dr}{d\theta}$, etc. But $\tan\phi=\frac{dy}{dx}$ which makes it something of $x$ and $y$. So having observed that we want to prove $\tan\psi=\frac{r}{dr/d\theta}$, which contains a derivative with respect to $\theta$, we use the identity $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}$ and $y=r\sin\theta$, $x=r\cos\theta$, then $\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}$. Then $$\tan\psi=\tan(\phi-\theta)\\=\frac{\tan\phi-\tan\theta}{1+\tan{\phi}\tan\theta}\\=\frac{\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}-\frac{\sin\theta}{\cos\theta}}{1+\frac{\frac{dr}{d\theta}\sin^2\theta+r\sin\theta\cos\theta}{\frac{dr}{d\theta}\cos^2\theta-r\sin\theta\cos\theta}}\\=\frac{\frac{\frac{dr}{d\theta}\sin\theta\cos\theta+r\cos^2\theta-\frac{dr}{d\theta}\sin\theta\cos\theta+r\sin^2\theta}{\frac{dr}{d\theta}\cos^2\theta-r\sin\theta\cos\theta}}{\frac{\frac{dr}{d\theta}}{\frac{dr}{d\theta}\cos^2\theta-r\sin\theta\cos\theta}}\\=\frac{r}{\frac{dr}{d\theta}}$$.