If in a triangle ABC, `(1 + cos A)/(a) + (1 + cos B)/(b) + (1+ cos C)/(c) = (k^(2) (1 + cos A) (1 + cos B) (1 + cos C))/(abc)`, then k is equal to

Options

A. `(1)/(2 sqrt2R)`
B. 2R
C. `(1)/(R)`
D. none of these

Explanation

To solve the given problem, we need to analyze the equation provided and find the value of \( k \).

### Given Equation:
\[ \frac{1 + \cos A}{a} + \frac{1 + \cos B}{b} + \frac{1 + \cos C}{c} = \frac{k^2 (1 + \cos A)(1 + \cos B)(1 + \cos C)}{abc} \]

### Step 1: Rewrite \( 1 + \cos A \), \( 1 + \cos B \), and \( 1 + \cos C \)
Using the identity:
\[ 1 + \cos A = 2 \cos^2\left(\frac{A}{2}\right) \]
We can similarly express \( 1 + \cos B \) and \( 1 + \cos C \):
\[ 1 + \cos B = 2 \cos^2\left(\frac{B}{2}\right), \quad 1 + \cos C = 2 \cos^2\left(\frac{C}{2}\right) \]

### Step 2: Express \( a, b, c \) in terms of \( R \) and \( \sin \)
Using the sine rule:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

### Step 3: Substitute into the Left-Hand Side (LHS)
Substituting the values into the LHS:
\[ \frac{2 \cos^2\left(\frac{A}{2}\right)}{2R \sin A} + \frac{2 \cos^2\left(\frac{B}{2}\right)}{2R \sin B} + \frac{2 \cos^2\left(\frac{C}{2}\right)}{2R \sin C} \]
This simplifies to:
\[ \frac{1}{R} \left( \frac{\cos^2\left(\frac{A}{2}\right)}{\sin A} + \frac{\cos^2\left(\frac{B}{2}\right)}{\sin B} + \frac{\cos^2\left(\frac{C}{2}\right)}{\sin C} \right) \]

### Step 4: Analyze the Right-Hand Side (RHS)
The RHS can be expressed as:
\[ \frac{k^2 (2 \cos^2\left(\frac{A}{2}\right))(2 \cos^2\left(\frac{B}{2}\right))(2 \cos^2\left(\frac{C}{2}\right))}{(2R \sin A)(2R \sin B)(2R \sin C)} \]
This simplifies to:
\[ \frac{k^2 \cdot 8 \cos^2\left(\frac{A}{2}\right) \cos^2\left(\frac{B}{2}\right) \cos^2\left(\frac{C}{2}\right)}{8R^3 \sin A \sin B \sin C} \]

### Step 5: Equate LHS and RHS
Setting the LHS equal to the RHS:
\[ \frac{1}{R} \left( \frac{\cos^2\left(\frac{A}{2}\right)}{\sin A} + \frac{\cos^2\left(\frac{B}{2}\right)}{\sin B} + \frac{\cos^2\left(\frac{C}{2}\right)}{\sin C} \right) = \frac{k^2 \cdot 8 \cos^2\left(\frac{A}{2}\right) \cos^2\left(\frac{B}{2}\right) \cos^2\left(\frac{C}{2}\right)}{8R^3 \sin A \sin B \sin C} \]

### Step 6: Solve for \( k^2 \)
From the equality, we can derive:
\[ 1 = k^2 \cdot \frac{R^2 \left( \frac{\cos^2\left(\frac{A}{2}\right)}{\sin A} + \frac{\cos^2\left(\frac{B}{2}\right)}{\sin B} + \frac{\cos^2\left(\frac{C}{2}\right)}{\sin C} \right)}{\cos^2\left(\frac{A}{2}\right) \cos^2\left(\frac{B}{2}\right) \cos^2\left(\frac{C}{2}\right)} \]

### Step 7: Find \( k \)
After simplifying and solving for \( k \), we find:
\[ k^2 = 4R^2 \implies k = 2R \]

### Final Answer:
Thus, the value of \( k \) is:
\[ \boxed{2R} \]