In any triangle ABC, the value of `(r_(1)+r_(2))/(1+cosC)` is equal to (where notation have their usual meaning) :

Options

A. 2R
B. 2r
C. R
D. `(2R^(2))/(r )`

Answer

To solve the problem, we need to find the value of \((r_1 + r_2) / (1 + \cos C)\) in triangle \(ABC\), where \(r_1\) and \(r_2\) are the ex-radii of the triangle. Let's go through the steps systematically.

### Step 1: Write the formulas for \(r_1\) and \(r_2\)
The ex-radii \(r_1\) and \(r_2\) are given by the following formulas:
\[ r_1 = \frac{4R \cdot \sin\left(\frac{A}{2}\right) \cdot \sin\left(\frac{B}{2}\right) \cdot \cos\left(\frac{C}{2}\right)}{1} \]
\[ r_2 = \frac{4R \cdot \sin\left(\frac{B}{2}\right) \cdot \sin\left(\frac{C}{2}\right) \cdot \cos\left(\frac{A}{2}\right)}{1} \]

### Step 2: Write the expression for \(1 + \cos C\)
Using the half-angle formula, we can express \(1 + \cos C\) as:
\[ 1 + \cos C = 1 + 2\cos^2\left(\frac{C}{2}\right) - 1 = 2\cos^2\left(\frac{C}{2}\right) \]

### Step 3: Substitute \(r_1\) and \(r_2\) into the expression
Now, we can substitute \(r_1\) and \(r_2\) into the expression \(\frac{r_1 + r_2}{1 + \cos C}\):
\[ \frac{r_1 + r_2}{1 + \cos C} = \frac{4R \left( \sin\left(\frac{A}{2}\right) \cdot \sin\left(\frac{B}{2}\right) \cdot \cos\left(\frac{C}{2}\right) + \sin\left(\frac{B}{2}\right) \cdot \sin\left(\frac{C}{2}\right) \cdot \cos\left(\frac{A}{2}\right) \right)}{2\cos^2\left(\frac{C}{2}\right)} \]

### Step 4: Simplify the expression
Factor out common terms:
\[ = \frac{2R \left( 2\sin\left(\frac{A}{2}\right) \cdot \sin\left(\frac{B}{2}\right) \cdot \cos\left(\frac{C}{2}\right) + 2\sin\left(\frac{B}{2}\right) \cdot \sin\left(\frac{C}{2}\right) \cdot \cos\left(\frac{A}{2}\right) \right)}{\cos^2\left(\frac{C}{2}\right)} \]

### Step 5: Use the sine addition formula
Using the sine addition formula:
\[ \sin\left(\frac{A + B}{2}\right) = \sin\left(\frac{A}{2}\right) \cdot \cos\left(\frac{B}{2}\right) + \cos\left(\frac{A}{2}\right) \cdot \sin\left(\frac{B}{2}\right) \]
We can express the numerator in terms of \(\sin\left(\frac{A + B}{2}\right)\):
\[ = \frac{2R \cdot \sin\left(\frac{A + B}{2}\right)}{\cos^2\left(\frac{C}{2}\right)} \]

### Step 6: Substitute \(A + B = \pi - C\)
Since \(A + B + C = \pi\), we have \(A + B = \pi - C\):
\[ = \frac{2R \cdot \sin\left(\frac{\pi - C}{2}\right)}{\cos^2\left(\frac{C}{2}\right)} \]
Using the identity \(\sin\left(\frac{\pi - C}{2}\right) = \cos\left(\frac{C}{2}\right)\):
\[ = \frac{2R \cdot \cos\left(\frac{C}{2}\right)}{\cos^2\left(\frac{C}{2}\right)} = 2R \]

### Final Answer
Thus, the value of \(\frac{r_1 + r_2}{1 + \cos C}\) is:
\[ \boxed{2R} \]