Certainly! Let's prove the given trigonometric identity:
[tex]\[
\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = \tan(A+B)
\][/tex]
### Step-by-Step Solution
1. Express Numerator using Difference of Squares:
The numerator is [tex]\( \sin^2 A - \sin^2 B \)[/tex]. We can use the difference of squares formula:
[tex]\[
\sin^2 A - \sin^2 B = (\sin A + \sin B)(\sin A - \sin B)
\][/tex]
2. Express Denominator using Angle Sum and Difference Identities:
Consider the denominator [tex]\( \sin A \cos A - \sin B \cos B \)[/tex].
We can rewrite each term using double-angle identities:
[tex]\[
\sin A \cos A = \frac{1}{2} \sin(2A) \quad \text{and} \quad \sin B \cos B = \frac{1}{2} \sin(2B)
\][/tex]
Therefore:
[tex]\[
\sin A \cos A - \sin B \cos B = \frac{1}{2} \sin (2A) - \frac{1}{2} \sin (2B) = \frac{1}{2} [\sin(2A) - \sin(2B)]
\][/tex]
3. Simplify the Original Expression:
Substituting the numerator and denominator expressions back into the given identity:
[tex]\[
\frac{(\sin A + \sin B)(\sin A - \sin B)}{\frac{1}{2} [\sin(2A) - \sin(2B)]}
\][/tex]
Simplify the fraction by multiplying the numerator and the denominator by 2:
[tex]\[
\frac{2 (\sin A + \sin B)(\sin A - \sin B)}{\sin(2A) - \sin(2B)}
\][/tex]
4. Use Sine Angle Sum and Difference Identities:
Now consider the sine angle sum and difference identities.
[tex]\[
\sin(2A) - \sin(2B) = 2 \cos\left(\frac{2A + 2B}{2}\right) \sin\left( \frac{2A - 2B}{2}\right)
\][/tex]
Simplifying further:
[tex]\[
\sin(2A) - \sin(2B) = 2 \cos(A + B) \sin(A - B)
\][/tex]
5. Transforming the Expression:
We substitute back this result:
[tex]\[
\frac{2 (\sin A + \sin B)(\sin A - \sin B)}{2 \cos(A+B) \sin(A-B)}
\][/tex]
The 2 in the numerator and denominator cancels out:
[tex]\[
\frac{(\sin A + \sin B)(\sin A - \sin B)}{\cos(A+B) \sin(A-B)}
\][/tex]
6. Recognize the Form of the Tangent Function:
Consider:
[tex]\[
(\sin A - \sin B) = 2 \cos\left( \frac{A+B}{2} \right) \sin\left( \frac{A-B}{2} \right)
\][/tex]
and
[tex]\[
(\sin A + \sin B) = 2 \sin\left( \frac{A+B}{2} \right) \cos\left( \frac{A-B}{2} \right)
\][/tex]
Substituting back, we note that:
[tex]\[
\frac{2 \sin\left( \frac{A+B}{2} \right) \cos\left( \frac{A-B}{2} \right) \cdot 2 \cos\left( \frac{A+B}{2} \right) \sin\left( \frac{A-B}{2} \right)}{ \cos(A+B) \sin(A-B)}
\][/tex]
[tex]\[
= \frac{4 \sin\left( \frac{A+B}{2} \right) \cos\left( \frac{A+B}{2} \right) \sin\left( \frac{A-B}{2} \right) \cos\left( \frac{A-B}{2} \right)}{ \cos(A+B) \sin(A-B)}
\][/tex]
Finally:
[tex]\[
= \frac{4 \frac{1}{2} \sin(A+B)}{\cos(A+B)}
= \tan(A+B)
\][/tex]
Therefore, we have proved:
[tex]\[
\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = \tan(A+B)
\][/tex]
