If the incircel of the triangle ABC, through it's circumcentre, then the `cos A + cos B + cos C` is

Options

A. ` -2`
B. `sqrt2`
C. `-sqrt2`
D. None of these

Answer

To solve the problem, we need to find the value of \( \cos A + \cos B + \cos C \) given that the incircle of triangle \( ABC \) passes through its circumcenter.

### Step-by-Step Solution:

1. **Understanding the Triangle and Circles**:
- Let \( I \) be the incenter of triangle \( ABC \) and \( O \) be the circumcenter.
- The incircle touches the sides of the triangle, and the circumcircle passes through the vertices of the triangle.

2. **Setting Up the Relationship**:
- The distance \( IO \) between the incenter and circumcenter can be expressed using the formula:
\[ IO^2 = R^2 - 2Rr \]
where \( R \) is the circumradius and \( r \) is the inradius.

3. **Given Condition**:
- Since the incircle passes through the circumcenter, we have \( IO = r \).
- Therefore, we can set up the equation:
\[ r^2 = R^2 - 2Rr \]

4. **Rearranging the Equation**:
- Rearranging gives us:
\[ R^2 + 2Rr - r^2 = 0 \]

5. **Dividing by \( R^2 \)**:
- Dividing the entire equation by \( R^2 \) yields:
\[ \left(\frac{r}{R}\right)^2 + 2\left(\frac{r}{R}\right) - 1 = 0 \]
- Let \( x = \frac{r}{R} \), then the equation becomes:
\[ x^2 + 2x - 1 = 0 \]

6. **Solving the Quadratic Equation**:
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2} \]

7. **Choosing the Positive Root**:
- Since \( \frac{r}{R} > 0 \), we take:
\[ \frac{r}{R} = \sqrt{2} - 1 \]

8. **Finding \( \cos A + \cos B + \cos C \)**:
- The formula for \( \cos A + \cos B + \cos C \) is given by:
\[ \cos A + \cos B + \cos C = 1 + \frac{r}{R} \]
- Substituting \( \frac{r}{R} = \sqrt{2} - 1 \):
\[ \cos A + \cos B + \cos C = 1 + (\sqrt{2} - 1) = \sqrt{2} \]

### Final Answer:
\[ \cos A + \cos B + \cos C = \sqrt{2} \]