In a triangle ABC, cos A+cos B+cos C=

Options

A. `1+r/R`
B. `1-r/R`
C. `1-R/r`
D. `1+R/r`

Answer

To solve the problem of finding the value of \( \cos A + \cos B + \cos C \) in a triangle \( ABC \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Triangle Properties**:
In any triangle \( ABC \), the angles \( A \), \( B \), and \( C \) satisfy the equation \( A + B + C = 180^\circ \).

2. **Using the Cosine Rule**:
We know that the cosine of angles in a triangle can be expressed in terms of the sides of the triangle. The cosine rule states:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

3. **Adding the Cosines**:
We will add the three cosine expressions:
\[ \cos A + \cos B + \cos C = \frac{b^2 + c^2 - a^2}{2bc} + \frac{a^2 + c^2 - b^2}{2ac} + \frac{a^2 + b^2 - c^2}{2ab} \]

4. **Finding a Common Denominator**:
The common denominator for the above fractions is \( 2abc \). We rewrite each term:
\[ \cos A + \cos B + \cos C = \frac{(b^2 + c^2 - a^2) \cdot a + (a^2 + c^2 - b^2) \cdot b + (a^2 + b^2 - c^2) \cdot c}{2abc} \]

5. **Simplifying the Numerator**:
Expanding the numerator:
\[ = \frac{ab^2 + ac^2 - a^3 + ba^2 + bc^2 - b^3 + ca^2 + cb^2 - c^3}{2abc} \]
Combine like terms.

6. **Using the Area of the Triangle**:
The area \( \Delta \) of triangle \( ABC \) can also be expressed using the sides and the semi-perimeter \( s \):
\[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]
where \( s = \frac{a+b+c}{2} \).

7. **Final Expression**:
After simplification, we find that:
\[ \cos A + \cos B + \cos C = 1 + \frac{r}{R} \]
where \( r \) is the inradius and \( R \) is the circumradius of triangle \( ABC \).

### Conclusion:
Thus, the value of \( \cos A + \cos B + \cos C \) in triangle \( ABC \) is given by:
\[ \cos A + \cos B + \cos C = 1 + \frac{r}{R} \]