`if cos A+cosB+cosC=1+4sin(A/2).sin(B/2).sin(C/2)` then `cos A+cosB+cosC=`

Options

A. `1-r/R`
B. `1+r/R`
C. `2-r/(2R)`
D. `2-(2r)/R`

Answer

To solve the equation \( \cos A + \cos B + \cos C = 1 + 4 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) \), we can use a known identity related to the circumradius and inradius of a triangle.

### Step-by-Step Solution:

1. **Understand the Given Identity**:
We start with the equation:
\[ \cos A + \cos B + \cos C = 1 + 4 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) \]

2. **Use the Inradius and Circumradius Relation**:
We know that for a triangle, there is a relation involving the inradius \( r \) and circumradius \( R \):
\[ r = 4R \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) \]
This means we can express \( 4 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) \) in terms of \( r \) and \( R \).

3. **Substituting the Relation**:
From the relation above, we can rewrite:
\[ 4 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) = \frac{r}{R} \]
Thus, substituting this back into our equation gives:
\[ \cos A + \cos B + \cos C = 1 + \frac{r}{R} \]

4. **Final Expression**:
Therefore, we can conclude:
\[ \cos A + \cos B + \cos C = 1 + \frac{r}{R} \]

### Conclusion:
Thus, the final expression for \( \cos A + \cos B + \cos C \) is:
\[ \cos A + \cos B + \cos C = 1 + \frac{r}{R} \]