Prove $ \cot \omega = \cot A + \cot B + \cot C$ condition given

Question

Question:

In the triangle $ABC$, lines $OA$, $OB$ and $OC$ are drawn so that the angles $OAB$, $OBC$ and $OCA$ are each equal to $\omega$, prove that

1. $ \cot\omega = \cot A + \cot B + \cot C$

2. $(\csc \omega)^2= (\csc A )^2+ (\csc B)^2 + (\csc C)^2$

My attempt:

Let point $O$ lie inside the triangle, then connect $O$ to $A,B$ and $C$.

Let $A=\omega+A'$

Let $B=\omega+B'$

Let $C=\omega+C'$

I tried to solve but not able to proceed from here onward.

Answer 1

I hope the following will help.

Since $OA$, $OB$ and $OC$ intersects in the common point $O$, we obtain: $$\sin^3\omega=\sin(\alpha-\omega)\sin(\beta-\omega)\sin(\gamma-\omega)$$ or $$\frac{1}{\prod\limits_{cyc}\sin\alpha}=\prod_{cyc}(\cot\omega-\cot\alpha),$$ which is a cubic equation of $\cot\omega$ and easy to show that $\sum\limits_{cyc}\cot\alpha$ is a root of this equation.

Indeed, we need to prove that $$\frac{1}{\prod\limits_{cyc}\sin\alpha}=\prod_{cyc}(\cot\alpha+\cot\beta)$$ or $$\frac{1}{\prod\limits_{cyc}\sin\alpha}=\prod_{cyc}\frac{\sin(\alpha+\beta)}{\sin\alpha\sin\beta},$$ which is obvious.

But I don't see what happens with two other roots of this equation.

The second follows from the first because $$1+\cot^2\alpha=\frac{1}{\sin^2\alpha}$$ and $$\sum_{cyc}\cot\alpha\cot\beta=1.$$