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.
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: