In `triangleABC`, the expression `(b^(2)-c^(2))/(asin(B-C)) + (c^(2)-a^2)/(bsin(C-A)) +(a^(2)-b^(2))/(csin(A-B))` is equal to

Options

A. `(3R)/2`
B. `3R`
C. R
D. 6R

Answer

To solve the expression

\[ \frac{b^2 - c^2}{a \sin(B - C)} + \frac{c^2 - a^2}{b \sin(C - A)} + \frac{a^2 - b^2}{c \sin(A - B)} \]

in triangle \(ABC\), we can utilize the relationship between the sides of the triangle and the angles, specifically using the Law of Sines which states:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]

where \(R\) is the circumradius of the triangle.

### Step 1: Substitute the sides in terms of \(R\)

Using the Law of Sines, we can express \(a\), \(b\), and \(c\) as follows:

\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

### Step 2: Rewrite the expression

Now we can substitute these values into the expression:

\[ \frac{(2R \sin B)^2 - (2R \sin C)^2}{(2R \sin A) \sin(B - C)} + \frac{(2R \sin C)^2 - (2R \sin A)^2}{(2R \sin B) \sin(C - A)} + \frac{(2R \sin A)^2 - (2R \sin B)^2}{(2R \sin C) \sin(A - B)} \]

### Step 3: Simplify each term

Each term simplifies as follows:

1. **First term:**
\[ \frac{4R^2 (\sin^2 B - \sin^2 C)}{2R \sin A \sin(B - C)} = \frac{2R (\sin^2 B - \sin^2 C)}{\sin A \sin(B - C)} \]

2. **Second term:**
\[ \frac{4R^2 (\sin^2 C - \sin^2 A)}{2R \sin B \sin(C - A)} = \frac{2R (\sin^2 C - \sin^2 A)}{\sin B \sin(C - A)} \]

3. **Third term:**
\[ \frac{4R^2 (\sin^2 A - \sin^2 B)}{2R \sin C \sin(A - B)} = \frac{2R (\sin^2 A - \sin^2 B)}{\sin C \sin(A - B)} \]

### Step 4: Combine the terms

Now we combine all three simplified terms:

\[ \frac{2R (\sin^2 B - \sin^2 C)}{\sin A \sin(B - C)} + \frac{2R (\sin^2 C - \sin^2 A)}{\sin B \sin(C - A)} + \frac{2R (\sin^2 A - \sin^2 B)}{\sin C \sin(A - B)} \]

### Step 5: Use trigonometric identities

Using the identity \( \sin^2 x - \sin^2 y = (\sin x + \sin y)(\sin x - \sin y) \), we can simplify further. However, we notice that each term will ultimately yield a common factor of \(2R\).

### Final Result

After simplification, we find that the entire expression equals:

\[ 6R \]

Thus, the final answer is:

\[ \boxed{6R} \]