In a triangle ABC, which of the following quantities denote the area of the triangles `(a^2-b^2)/2(sinA sinB)/(sin(A-B))`

Answer

To determine whether the quantity \(\frac{a^2 - b^2}{2 \sin A \sin B \sin(A - B)}\) denotes the area of triangle ABC, we can follow these steps:

### Step 1: Start with the given expression
The expression we need to analyze is:
\[ \frac{a^2 - b^2}{2 \sin A \sin B \sin(A - B)} \]

### Step 2: Use the Law of Sines
According to the Law of Sines, we know that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
where \(R\) is the circumradius of the triangle. From this, we can express \(a\) and \(b\) in terms of \(R\):
\[ a = 2R \sin A \quad \text{and} \quad b = 2R \sin B \]

### Step 3: Substitute \(a\) and \(b\) in the expression
Substituting these values into the expression gives:
\[ a^2 = (2R \sin A)^2 = 4R^2 \sin^2 A \]
\[ b^2 = (2R \sin B)^2 = 4R^2 \sin^2 B \]
Thus, we can rewrite the expression:
\[ \frac{4R^2 \sin^2 A - 4R^2 \sin^2 B}{2 \sin A \sin B \sin(A - B)} \]

### Step 4: Factor out common terms
Factoring out \(4R^2\) from the numerator:
\[ \frac{4R^2 (\sin^2 A - \sin^2 B)}{2 \sin A \sin B \sin(A - B)} \]

### Step 5: Simplify the expression
Using the identity \(\sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B)\):
\[ \frac{4R^2 (\sin A - \sin B)(\sin A + \sin B)}{2 \sin A \sin B \sin(A - B)} \]
Now, simplifying gives:
\[ \frac{2R^2 (\sin A - \sin B)(\sin A + \sin B)}{\sin A \sin B \sin(A - B)} \]

### Step 6: Use the sine addition formula
Using the sine addition formula, we know:
\[ \sin(A + B) = \sin C \]
Thus, we can express \(\sin(A + B)\) in terms of \(C\):
\[ \sin(A + B) = \sin(\pi - C) = \sin C \]

### Step 7: Substitute back and find the area
From the properties of triangles, the area \(K\) can be expressed as:
\[ K = \frac{1}{2}ab \sin C \]
Substituting \(a = 2R \sin A\) and \(b = 2R \sin B\):
\[ K = \frac{1}{2} (2R \sin A)(2R \sin B) \sin C = 2R^2 \sin A \sin B \sin C \]

### Conclusion
After simplification, we find that the original expression indeed represents the area of triangle ABC:
\[ \frac{a^2 - b^2}{2 \sin A \sin B \sin(A - B)} = \frac{1}{2} ab \sin C \]

Thus, the quantity \(\frac{a^2 - b^2}{2 \sin A \sin B \sin(A - B)}\) denotes the area of triangle ABC.

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