In a triangle ABC if `r_1=R`, then

Options

A. `cos A+ cos B = cos C`
B. `cos B+ cos C = cos A`
C. `cos C+ cos A = cos B`
D. `cos A+ cos B + cos C = 1`

Answer

To solve the problem where \( r_1 = R \) in triangle \( ABC \), we need to understand the relationship between the inradius \( r \) and circumradius \( R \) of a triangle.

### Step-by-step Solution:

1. **Understanding the Notations**:
- Let \( R \) be the circumradius of triangle \( ABC \).
- Let \( r_1 \) be the inradius of triangle \( ABC \).
- The relationship given is \( r_1 = R \).

2. **Using the Formula for Inradius and Circumradius**:
- The inradius \( r \) can be expressed in terms of the area \( A \) and the semi-perimeter \( s \) of the triangle:
\[ r = \frac{A}{s} \]
- The circumradius \( R \) can be expressed in terms of the sides \( a, b, c \) of the triangle and the area \( A \):
\[ R = \frac{abc}{4A} \]

3. **Equating Inradius and Circumradius**:
- Given \( r_1 = R \), we can set up the equation:
\[ \frac{A}{s} = \frac{abc}{4A} \]

4. **Cross-Multiplying**:
- Cross-multiply to eliminate the fractions:
\[ A^2 = \frac{abc \cdot s}{4} \]

5. **Using the Semi-perimeter**:
- The semi-perimeter \( s \) is defined as:
\[ s = \frac{a + b + c}{2} \]
- Substitute \( s \) into the equation:
\[ A^2 = \frac{abc \cdot \frac{a + b + c}{2}}{4} \]

6. **Simplifying**:
- This simplifies to:
\[ A^2 = \frac{abc(a + b + c)}{8} \]

7. **Conclusion**:
- The condition \( r_1 = R \) implies a specific relationship between the sides of the triangle and its area. This is not a common case and typically occurs in special triangles, such as equilateral triangles.