In the given triangle we have this point $O$ such that $\angle OAB=\angle OBC=\angle OCA=\omega$ Hence prove that $\cot\omega=\cot A+\cot B+\cot C$.
I figured out the RHS by using sine and cosine identities but the LHS couldn't be worked out by me. Please help.
We are dealing with the Brocard angle. Notice that, by the trigonometric form of Ceva's theorem,
$$ \sin(\omega)^3 = \sin(A-\omega)\sin(B-\omega)\sin(C-\omega). \tag{1}$$ Use the sine addition formulas to write the RHS of $(1)$ as a sum of sines, then write $\sin(A)$ as $\frac{2\Delta}{bc}$ and $\cos(A)$ as $\frac{b^2+c^2-a^2}{2bc}$. Do the same for $B$ and $C$ and that will lead to: $$ \cot(\omega) = \frac{a^2+b^2+c^2}{4\Delta}=\cot(A)+\cot(B)+\cot(C).\tag{2} $$
You may prove the same by proving first that the trilinear coordinates of the Brocard points are $$\left[\frac{b}{c};\frac{c}{a};\frac{a}{b}\right]\quad\text{and}\quad \left[\frac{c}{b};\frac{a}{c};\frac{b}{a}\right]. \tag{3}$$
