<s>[INST] Geometry Rotation and Trigonometry 
$A, B, C$ and $D$ are on a line such that $AB=BC=CD$. Also, $P$ is a point on a circle with $BC$ as a diameter. Find $\tan\angle{APB} \cdot \tan\angle{CPD}$.

Let $O$ be the center of $(BPC)$. Let $P$' be the point of intersection of $PO$ and $(BOC)$ again. Then $C$ is the centroid of $PP'D$ since $OD$ is a median and $CD=2OD$. By symmetry,
$$\tan\angle{APB} \cdot \tan\angle{CPD}=\tan\angle{CP'D} \cdot \tan\angle{CPD}$$
This all looks promising and hopefully, it will help (or maybe it is misleading). 
Thanks!
 [/INST] 
Let $A\equiv(0,0),B\equiv(1,0),C\equiv(2,0)$ and $D\equiv(3,0)$. Also let $\angle PBC=\theta$
$PB=\cos \theta$. Also, slope of $BP=\tan\theta$.
So, co-ordinates of $P\equiv(1+\cos^2\theta,\sin\theta \cos\theta)$
So, slope of $AP=\frac{\tan\theta}{2+\tan^2\theta}$
$tan\alpha=\frac{\tan\theta-\frac{\tan\theta}{2+\tan^2\theta}}{1+\frac{tan^2\theta}{2+\tan^2\theta}}=\frac{\tan\theta}{2}$
Slope of $CP=-\cot\theta$
$$Slope(PD)=\frac{\sin\theta\cos\theta}{\cos^2\theta-2}=-\frac{\tan\theta}{1+2\tan^2\theta}$$
similarly $\tan\beta=\frac{1}{2\tan\theta}$
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