`(sin^2A-sin^2B)/(sinAcosA-sinBcosB)` is equal to
(a) `tan(A-B)`
(b) `"tan"(A+B)`
(c) `cot(A-B)`
(d) `cot(A+B)`

Options

A. `tan(A-B)`
B. `tan(A+B)`
C. `cot(A-B)`
D. `cot(A+B)`

Explanation

To solve the expression \(\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B}\), we can follow these steps:

### Step 1: Use the difference of squares
The numerator \(\sin^2 A - \sin^2 B\) can be factored using the difference of squares formula:
\[ \sin^2 A - \sin^2 B = (\sin A + \sin B)(\sin A - \sin B) \]

### Step 2: Rewrite the denominator
The denominator \(\sin A \cos A - \sin B \cos B\) can be rewritten as:
\[ \sin A \cos A - \sin B \cos B = \frac{1}{2} \sin(2A) - \frac{1}{2} \sin(2B) = \frac{1}{2} (\sin(2A) - \sin(2B)) \]
Using the sine difference formula, we have:
\[ \sin(2A) - \sin(2B) = 2 \cos(A + B) \sin(A - B) \]
Thus, the denominator becomes:
\[ \sin A \cos A - \sin B \cos B = \frac{1}{2} \cdot 2 \cos(A + B) \sin(A - B) = \cos(A + B) \sin(A - B) \]

### Step 3: Substitute back into the expression
Now substituting the factored forms back into the original expression, we have:
\[ \frac{(\sin A + \sin B)(\sin A - \sin B)}{\cos(A + B) \sin(A - B)} \]

### Step 4: Simplify the expression
Notice that \(\sin A - \sin B\) cancels out:
\[ = \frac{\sin A + \sin B}{\cos(A + B)} \]

### Step 5: Use the identity for tangent
Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we can express the result as:
\[ = \tan(A + B) \]

### Final Result
Thus, the expression \(\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B}\) simplifies to:
\[ \tan(A + B) \]

### Answer
The correct option is (b) \(\tan(A + B)\).

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