In the triangle ABC, lines `OA,OB` and `OC` are drawn so that angles OAB, OBC and OCA are each equal to `omega`, prove that `cotomega=cotA+cotB+cotC`

Answer

To prove that in triangle ABC, if angles OAB, OBC, and OCA are each equal to ω, then we have:

\[ \cot(\omega) = \cot(A) + \cot(B) + \cot(C) \]

### Step-by-Step Solution:

1. **Identify the Angles and Sides**:
- Let the angles of triangle ABC be A, B, and C.
- Let the sides opposite to these angles be denoted as a, b, and c respectively.

2. **Apply the Sine Rule**:
- According to the sine rule in triangle ABC:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = \lambda \quad (\text{a constant}) \]

3. **Analyze Triangle AOB**:
- In triangle AOB, the angle OAB is ω, and the angle OBA is \(B - \omega\).
- The angle AOB can be expressed as:
\[ AOB = \pi - (OAB + OBA) = \pi - (\omega + (B - \omega)) = \pi - B \]
- By the sine rule in triangle AOB:
\[ \frac{AO}{\sin(B - \omega)} = \frac{c}{\sin(\pi - B)} = \frac{c}{\sin B} \]

4. **Set Up the Equation**:
- From the sine rule in triangle AOB:
\[ AO = \frac{c \cdot \sin(B - \omega)}{\sin B} \]

5. **Analyze Triangle ACO**:
- In triangle ACO, the angle OAC is ω, and the angle OCA is \(A - \omega\).
- The angle AOC can be expressed as:
\[ AOC = \pi - (OAC + OCA) = \pi - (\omega + (A - \omega)) = \pi - A \]
- By the sine rule in triangle ACO:
\[ \frac{AO}{\sin(A - \omega)} = \frac{b}{\sin(\pi - A)} = \frac{b}{\sin A} \]

6. **Set Up the Second Equation**:
- From the sine rule in triangle ACO:
\[ AO = \frac{b \cdot \sin(A - \omega)}{\sin A} \]

7. **Equate the Two Expressions for AO**:
- Set the two expressions for AO equal:
\[ \frac{c \cdot \sin(B - \omega)}{\sin B} = \frac{b \cdot \sin(A - \omega)}{\sin A} \]

8. **Use Sine Difference Formula**:
- Use the sine difference formula:
\[ \sin(B - \omega) = \sin B \cos \omega - \cos B \sin \omega \]
- Substitute this into the equation:
\[ \frac{c (\sin B \cos \omega - \cos B \sin \omega)}{\sin B} = \frac{b \cdot \sin(A - \omega)}{\sin A} \]

9. **Simplify and Rearrange**:
- Rearranging gives:
\[ c \cos \omega - c \frac{\cos B \sin \omega}{\sin B} = \frac{b \cdot \sin(A - \omega)}{\sin A} \]

10. **Repeat for Other Triangles**:
- Repeat similar steps for triangles BOC and AOC to derive similar equations.

11. **Combine Results**:
- Combine the results from all triangles to arrive at:
\[ \cot(\omega) = \cot(A) + \cot(B) + \cot(C) \]

### Final Result:
Thus, we have proven that:
\[ \cot(\omega) = \cot(A) + \cot(B) + \cot(C) \]