If R denotes circumradius, then in `DeltaABC,(b^2-c^2)/(2aR)` is equal to

Answer

To solve the problem, we need to find the value of \(\frac{b^2 - c^2}{2aR}\) in triangle \(ABC\), where \(R\) is the circumradius.

### Step-by-Step Solution:

1. **Understanding the Circumradius**:
The circumradius \(R\) of triangle \(ABC\) can be expressed in terms of the sides and angles of the triangle. We know that:
\[ R = \frac{a}{2 \sin A} \]

2. **Using the Sine Rule**:
According to the sine rule, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
From this, we can express the sides in terms of \(k\):
\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]

3. **Substituting for \(R\)**:
We substitute \(R\) into our expression:
\[ \frac{b^2 - c^2}{2aR} = \frac{b^2 - c^2}{2a \cdot \frac{a}{2 \sin A}} = \frac{b^2 - c^2}{\frac{a^2}{\sin A}} = \frac{(b^2 - c^2) \sin A}{a^2} \]

4. **Expressing \(b^2 - c^2\)**:
We can use the identity \(b^2 - c^2 = (b - c)(b + c)\). Thus, we need to express \(b\) and \(c\) in terms of \(k\):
\[ b = k \sin B, \quad c = k \sin C \]
Therefore:
\[ b^2 - c^2 = k^2 (\sin^2 B - \sin^2 C) \]

5. **Using the Sine Difference Identity**:
We can apply the sine difference identity:
\[ \sin^2 B - \sin^2 C = (\sin B - \sin C)(\sin B + \sin C) \]

6. **Substituting Back**:
Now substituting back into our expression:
\[ \frac{(b^2 - c^2) \sin A}{a^2} = \frac{k^2 (\sin B - \sin C)(\sin B + \sin C) \sin A}{a^2} \]

7. **Final Expression**:
Since \(a = k \sin A\), we can express \(a^2\) as:
\[ a^2 = k^2 \sin^2 A \]
Thus, we have:
\[ \frac{(b^2 - c^2) \sin A}{a^2} = \frac{(\sin B - \sin C)(\sin B + \sin C) \sin A}{\sin^2 A} \]

8. **Result**:
Finally, using the sine difference and sum identities, we find:
\[ \frac{b^2 - c^2}{2aR} = \sin(B - C) \]

### Conclusion:
Thus, we conclude that:
\[ \frac{b^2 - c^2}{2aR} = \sin(B - C) \]