Brocard Angles proof by Sine and cosine formulae. 
The angles denoted by $\omega$ are the Brocard angles. Recently i came to know about the Brocard Angles and also their property i.e $\cot{\omega}=\cot{A}+\cot{B}+\cot{C}$. In my previous question I got the answer on proving the identity.  But I want to prove this identity only through the sines and cosine formula(excluding any excessive use of geometry parts)
Now I tried the question this way:

In $\triangle APC$, $\angle CAP=A-\omega; \angle CPA=\pi-A$.
Thus using the sine rule we can write $\frac{\sin{(A-\omega)}}{CP} = \frac{\sin{A}}{b}$.
Similarly using the same rule in other triangle we can write:
For $\triangle CPB$ $\frac{\sin{(C-\omega)}}{PB} = \frac{\sin{C}}{a}$
For $\triangle APB$ $\frac{\sin{(B-\omega)}}{AP} = \frac{\sin{B}}{c}$
Now in the respective sine formulae I expanded the expressions of $\sin{(A-\omega)}$, $\sin{(B-\omega)}$ and $\sin{(C-\omega)}$ 
This gave me-
  $$\frac{CP}{b}=\cos{\omega}-\sin{\omega}\cot{A}\tag{1}$$
  $$\frac{PB}{a}=\cos{\omega}-\sin{\omega}\cot{C}\tag{2}$$
  $$\frac{AP}{c}=\cos{\omega}-\sin{\omega}\cot{B}\tag{3}$$
Adding the equations $(1),(2)$ and $(3)$
  $$\frac{CP}{b}+\frac{PB}{a}+\frac{AP}{c}=3\cos{\omega}-\sin{\omega}(\cot{A}+\cot{B}+\cot{C})$$

I got stuck after this. Please tell me whether I can proceed further or is my method completely inconclusive.
 A: Hint:
$$\cot{\omega}=\cot{A}+\cot{B}+\cot{C} \Longleftrightarrow \sin{(A-\omega)}\sin{(B-\omega)}\sin{(C-\omega)}=\sin^3{\omega}\tag{1}$$
Now using sine rule, we get
$$\dfrac{\sin{(A-\omega)}}{\sin{\omega}}=\frac{CP}{AP}\tag{2}$$
$$\dfrac{\sin{(B-\omega)}}{\sin{\omega}}=\frac{AP}{BP}\tag{3}$$
$$\dfrac{\sin{(C-\omega)}}{\sin{\omega}}=\frac{BP}{CP}\tag{4}$$
Multiplying $(2),(3)$ and $(4)$ out, we get $(1)$

Proof of Hint:

 $$\quad \quad \cot{\omega}=\cot{A}+\cot{B}+\cot{C}$$
 $$\Longleftrightarrow \cot{\omega}-\cot{A}=\cot{B}+\cot{C}$$
 $$\Longleftrightarrow \dfrac{\sin{(A-\omega})}{\sin{A}\sin{\omega}}=\dfrac{\sin{(B+C)}}{\sin{B}\sin{C}}=\dfrac{\sin{(\pi-A)}}{\sin{B}\sin{C}}=\dfrac{\sin{A}}{\sin{B}\sin{C}}$$
 So, $$\sin{(A-\omega)}=\dfrac{\sin^2{A}\sin{\omega}}{\sin{B}\sin{C}}\tag{5}$$
 Similarly, $$\sin{(B-\omega)}=\dfrac{\sin^2{B}\sin{\omega}}{\sin{A}\sin{C}}\tag{6}$$
 and $$\sin{(C-\omega)}=\dfrac{\sin^2{C}\sin{\omega}}{\sin{A}\sin{B}}\tag{7}$$
 Multiplying $(5),(6)$ and $(7)$ out, we get $(1)$.

A: Here I also got one more thing, this can be proved in one more way.
In the figure 
We can see that $$\frac{\sin(A-\omega)}{PC}=\frac{\sin \angle CPA}{b}$$
Here through basic geometry we can easily conclude that $\angle CPA=\pi - A$, hence the above equation can be written as:
$$\frac{\sin(A-\omega)}{PC}=\frac{\sin A}{b}$$
Also in $\triangle PCB$ we can use sine rule
$$\frac{\sin(\omega)}{PC}=\frac{\sin( \angle CPB)}{a}\tag1$$
Here also we can conclude that $\angle CPB=\pi -C$
Hence $$\frac{\sin(\omega)}{PC}=\frac{\sin(C)}{a}\tag2$$
Now dividing equation $(1)$ by equation $(2)$
$$\frac{\sin (A-\omega)}{\sin(\omega)}=\frac{a \sin A}{b \sin C}$$
Using sine rule we can also write:
$$\frac{\sin (A-\omega)}{\sin(\omega)}=\frac{\lambda \sin A \cdot \sin A}{\lambda \sin B \cdot \sin C}$$
$$\frac{\sin (A-\omega)}{\sin(\omega)}=\frac{ \sin A \cdot \sin (\pi -(B+C)}{ \sin B \cdot \sin C}$$
$$\frac{\sin A \cos \omega- \cos A \cdot \sin \omega}{sin(\omega)}=\frac{ \sin A \cdot (\sin B \cdot \cos C + \sin C \cdot \cos B)}{ \sin B \cdot \sin C}$$
$${\sin A \cot \omega- \cos A }={ \sin A \cdot ( \cot C + \cot B)}$$
Dividing both sides $\sin A$
$$\cot\omega = \cot A + \cot B + \cot C $$
