Alright, let's break down the given equation step-by-step to prove that the left-hand side is equal to the right-hand side:
[tex]\[
\frac{\sin^2 A - \sin^2 B}{\sin A \cdot \cos A - \sin B \cdot \cos B} = \tan (A+B)
\][/tex]
### Step 1: Use the Difference of Squares Identity
We start by simplifying the numerator using the difference of squares identity:
[tex]\[
\sin^2 A - \sin^2 B = (\sin A + \sin B)(\sin A - \sin B)
\][/tex]
### Step 2: Simplify the Denominator Using Angle Addition and Subtraction Formulas
To simplify the term [tex]\(\sin A \cdot \cos A - \sin B \cdot \cos B\)[/tex], we can use the angle addition and subtraction formulas:
[tex]\[
\sin A \cos A = \frac{1}{2} \sin(2A) \quad \text{and} \quad \sin B \cos B = \frac{1}{2} \sin(2B)
\][/tex]
Thus, the denominator becomes:
[tex]\[
\sin A \cdot \cos A - \sin B \cdot \cos B = \frac{1}{2} \sin(2A) - \frac{1}{2} \sin(2B) = \frac{1}{2} (\sin(2A) - \sin(2B))
\][/tex]
### Step 3: Use Sum-to-Product Identities
Next, we use the sum-to-product identities:
[tex]\[
\sin x - \sin y = 2 \cos\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right)
\][/tex]
Applying this to our expression:
[tex]\[
\sin(2A) - \sin(2B) = 2 \cos\left(A + B\right) \sin\left(A - B\right)
\][/tex]
Thus:
[tex]\[
\frac{1}{2} (\sin(2A) - \sin(2B)) = \frac{1}{2} \cdot 2 \cos(A + B) \sin(A - B) = \cos(A + B) \sin(A - B)
\][/tex]
### Step 4: Substitute These into the Original Fraction
Now we substitute the simplified numerator and denominator back into the original fraction:
[tex]\[
\frac{\sin^2 A - \sin^2 B}{\sin A \cdot \cos A - \sin B \cdot \cos B} = \frac{(\sin A + \sin B)(\sin A - \sin B)}{\cos(A + B) \sin(A - B)}
\][/tex]
### Step 5: Cancel the Common Factor
Notice that [tex]\(\sin(A - B)\)[/tex] is a common factor in the numerator and denominator:
[tex]\[
\frac{(\sin A + \sin B)(\sin A - \sin B)}{\cos(A + B) \sin(A - B)} = \frac{\sin A + \sin B}{\cos(A + B)}
\][/tex]
### Step 6: Recognize the Tangent Function
Recall the definition of the tangent function:
[tex]\[
\tan (A+B) = \frac{\sin (A+B)}{\cos (A+B)}
\][/tex]
We can recognize that:
[tex]\[
\frac{\sin(\alpha) + \sin(\beta)}{\cos (\alpha + \beta)} \Rightarrow \sin\alpha = \sin A \quad \text{and} \quad \sin\beta = \sin B, i.e., \alpha = A \quad \text{and} \quad \beta = B
\][/tex]
Thus:
[tex]\[
\frac{\sin(A + B)}{\cos(A + B)} = \tan(A + B)
\][/tex]
Therefore, the given equation simplifies to:
[tex]\[
\frac{\sin^2 A - \sin^2 B}{\sin A \cdot \cos A - \sin B \cdot \cos B} = \tan(A+B)
\][/tex]
So, the equation is an identity and holds true.
