Help with Polar Coordinates proof in Calculus so I have this Polar Coordinates Homework Problem, and I don't know really know how to start/prove it. The problem goes as follows:
Let $f$ be a differentiable function, and let $P$ be the point $(\theta,f(\theta))$ on the polar graph of $r = f(\theta).$ Let $\alpha$ be the angle between the tangent line to the graph at $P,$ and the line $OP,$ where $O$ is the origin. Assuming that $f'(\theta) \neq 0,$ show that
$\tan\alpha=\frac{f(\theta)}{f'(\theta)}$

From what I've understood from this problem, if I extended a tangent line to P all the way to the x-axis, and called that point Q, I make $\angle PQO=\pi-\theta-\alpha,$ beyond that, I am lost.
 A: Hint:
Firstly, we have: $$x=r\cos \theta$$ and $$y=r\sin \theta $$
Then, we have, $$\frac {dy}{dx}=\frac{\frac {dy}{d\theta}}{\frac {dx}{d\theta}}$$
Be careful to note that $r=f(\theta)$.
Now, slope of tangent is known. Also, slope of line $OP$ is simply $\tan \theta$. Can you find out the angle between two lines, if the slopes of both lines are known?
Note: With respect to a coordinate system, if slope of two lines is given by $m_1$ and $m_2$, then the angle between the two lines, $\alpha$, is given by-
$$\tan \alpha=\frac {|m_2-m_1|}{|1+m_1m_2|}$$
A: It goes direct, fast and advantageous to consider differentials as segments in a right triangle of differentials.. permitted by basic consideration of first order differential definition in geometry.
$$ \tan \alpha =\frac {AB}{AP} =\frac{r d \theta }{dr}  = \frac{r}{dr/d \theta}=\frac{f(\theta) }{f'(\theta) }$$

A: The slope of the tangent line is $ \dfrac{dy}{dx} $ and we have
$ \dfrac{dy}{dy} = \dfrac{ \dfrac{dy}{d \theta} }{ \dfrac{dx}{d \theta}} $
Now, $ x = f(\theta) \cos \theta $ , so
$ \dfrac{dx}{d \theta} = f'(\theta) \cos \theta - f(\theta) \sin \theta $
and $ y = f(\theta) \sin \theta $, therefore,
$ \dfrac{ dy } {d \theta } = f'(\theta) \sin \theta + f(\theta) \cos \theta $
Hence,
$ \dfrac{dy}{dx} = \dfrac{f'(\theta) \sin \theta + f(\theta) \cos \theta }{f'(\theta) \cos \theta - f(\theta) \sin \theta} $
Dividing by $ \cos \theta $ top and bottom,
$ m_1 =  \dfrac{dy}{dx} = \dfrac{ f'(\theta) \tan \theta + f( \theta) } { f'(\theta) - f(\theta) \tan \theta } $
This is the slope of the tangent line, and $m_2 = \tan \theta $ is the slope of $OP$
Hence, the tangent of the angle between the two lines is given by
$\tan \alpha = \dfrac{m_1-m_2}{1 + m_1 m_2 } $
$ \tan \alpha = \dfrac{ f'(\theta) m_2 + f(\theta) - m_2 (f'(\theta) - m_2 f(\theta) } { (f'(\theta) - m_2 f(\theta)) + m_2 (f'(\theta) m_2 + f(\theta) ) }$
Simplifying,
$ \tan \alpha = \dfrac{ (1 + m_2^2) f(\theta) }{(1 + m_2^2) f'(\theta)} = \dfrac{f(\theta)}{f'(\theta) } $
A: We can use the unit vectors $$\hat r=\cos{\theta}\ \hat i + \sin\theta\ \hat j$$ and $$\hat \theta=-\sin{\theta}\ \hat i + \cos\theta\ \hat j$$ to express $f$ and $f'$ as vector equations. Note that $\hat\theta=\frac{d\hat r}{d\theta}$
We write $$\vec r=r\hat r=f(\theta)\hat r$$ and $$\frac{d\vec r}{d\theta}=f'(\theta)\hat r+f(\theta)\frac{d\hat r}{d\theta}=f'(\theta)\hat r + f(\theta)\hat\theta$$
We can divide the magnitude of the cross product by the dot product to get the tangent of the angle between them.
$$\tan{\alpha}=\frac{|\vec r\times\frac{d\vec r}{d\theta}|}{\vec r\cdot\frac{d\vec r}{d\theta}}=\frac{f(\theta)^2}{f(\theta)f'(\theta)}=\frac{f(\theta)}{f'(\theta)}$$
