`(sin(B-C))/(sin(B+C))=`

Options

A. `(c^2)/(a^2-b^2)`
B. `(b^2)/(c^2-a^2)`
C. `(a^2)/(b^2-c^2)`
D. `(b^2+c^2)/(a^2)`

Answer

To solve the expression \(\frac{\sin(B-C)}{\sin(B+C)}\), we can use trigonometric identities and properties of triangles. Here's a step-by-step solution:

### Step 1: Use the Sine Addition and Subtraction Formulas
We start by applying the sine addition and subtraction formulas:

\[ \sin(B - C) = \sin B \cos C - \cos B \sin C \]
\[ \sin(B + C) = \sin B \cos C + \cos B \sin C \]

### Step 2: Substitute the Formulas into the Expression
Now, substituting these formulas into our expression gives:

\[ \frac{\sin(B-C)}{\sin(B+C)} = \frac{\sin B \cos C - \cos B \sin C}{\sin B \cos C + \cos B \sin C} \]

### Step 3: Simplify the Expression
Let’s denote \(x = \sin B \cos C\) and \(y = \cos B \sin C\). The expression simplifies to:

\[ \frac{x - y}{x + y} \]

### Step 4: Use the Sine Rule
In a triangle, we can apply the sine rule, which states:

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

From this, we can express \(\sin B\) and \(\sin C\) in terms of the sides of the triangle:

\[ \sin B = k \cdot b, \quad \sin C = k \cdot c \]

### Step 5: Substitute Back into the Expression
Substituting these into our expression:

\[ \sin B \cos C = (k \cdot b) \cos C, \quad \cos B \sin C = \cos B (k \cdot c) \]

### Step 6: Use the Cosine Rule
Using the cosine rule, we can express \(\cos B\) in terms of the sides:

\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

### Step 7: Substitute and Simplify
Now substitute \(\cos B\) back into our expression. After substituting and simplifying, we will reach a point where we can cancel terms and simplify further.

### Step 8: Final Simplification
After careful simplification, we will arrive at:

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

### Conclusion
Thus, the final result is:

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