Relation of tangent Right triangle 
From The picture
$$\frac{tan\varphi}{tan\psi}=\frac{AC}{BC}$$
Question:
Prove these relations
 A: HINT....Draw a line through $E$ parallel to $AB$ and write down expressions for $\tan\theta, \tan\phi$ and so on. The first result follows immediately. The second result is another way of expressing the relationship between sides of various similar triangles. I hope this helps.
A: I assume that $\angle{CAD}=\angle{CBE}=90^\circ$.
HINT : 
We can prove the two equations using the followings : 
$$\tan\theta=\frac{AD}{AC},\quad \tan\varphi=\frac{AD}{AB},\quad\tan\psi=\frac{BE}{AB}\tag1$$
Also, since $\triangle{CAD}$ is similar to $\triangle{CBE}$,
$$AC:BC=AD:BE\tag2$$
A: Let $AC=d$ then we have 
In right $\triangle CAD$ $$\tan \theta=\frac{AD}{AC}\iff AD=d\tan \theta\tag 1$$
In right $\triangle BAD$ $$\tan \phi=\frac{AD}{AB}\iff AB=AD\cot \phi=d\tan \theta\cot \phi\tag 2$$
In right $\triangle ABE$ $$\tan \psi=\frac{BE}{AB}\iff BE=d\tan \theta\tan \psi\cot \phi\tag 3$$
In right $\triangle CBE$ $$\tan \theta=\frac{BE}{BC}\iff BC=d\tan \psi\cot \phi\tag 4$$
1. from (2), we have $$\tan \phi=\frac{AD}{AB}$$
Setting value of $AD$ from (1)
$$\tan \phi=\frac{AC\tan \theta}{AB}$$
$$\tan \phi=\tan \theta\left(\frac{AB+BC}{AB}\right)$$
$$\tan \phi=\tan \theta+\tan \theta \left(\frac{BC}{AB}\right)$$
Setting value of $\tan \theta$ from (4)
$$\tan \phi=\tan \theta+\frac{BE}{BC} \left(\frac{BC}{AB}\right)$$
$$\tan \phi=\tan \theta+\frac{BE}{AB} $$
Setting value of $\frac{BE}{AB}$ from (3)
$$\color{red}{\tan \phi}=\color{blue}{\tan \theta+\tan \psi}$$


*In similar triangles $\triangle DAC$ & $\triangle EBC$, we have 
$$\frac{AC}{BC}=\frac{AD}{BE}$$
Setting the values of $AD$ & $BE$ from (1)& (3) respectively, we get 
$$\frac{AC}{BC}=\frac{d\tan \theta}{d\tan \theta\tan \psi\cot \phi}$$
$$\color{red}{\frac{AC}{BC}}=\color{blue}{\frac{\tan \phi}{\tan \psi}}$$

