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

Explanation

To solve the expression \((\sin^2 A - \sin^2 B) / (\sin A \cos A - \sin B \cos B)\), we will follow these steps:

### Step 1: Multiply and Divide by 2
We start by multiplying and dividing the expression by 2:
\[ \frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = \frac{2(\sin^2 A - \sin^2 B)}{2(\sin A \cos A - \sin B \cos B)} \]

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