Prove that: \[ \frac{\sin A+\sin B}{\sin A-\sin B} = \tan\frac{A+B}{2} \cot\frac{A-B}{2} \]
Solution
Consider the left-hand side:
\[
\frac{\sin A+\sin B}{\sin A-\sin B}
\]
Using the identities:
\[
\sin A+\sin B
=
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
\]
\[
\sin A-\sin B
=
2\cos\frac{A+B}{2}\sin\frac{A-B}{2}
\]
Substituting these values:
\[
=
\frac{
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
}{
2\cos\frac{A+B}{2}\sin\frac{A-B}{2}
}
\]
Cancelling common terms:
\[
=
\frac{
\sin\frac{A+B}{2}
}{
\cos\frac{A+B}{2}
}
\times
\frac{
\cos\frac{A-B}{2}
}{
\sin\frac{A-B}{2}
}
\]
\[
=
\tan\frac{A+B}{2}
\cot\frac{A-B}{2}
\]
Hence,
\[
\boxed{
\frac{\sin A+\sin B}{\sin A-\sin B}
=
\tan\frac{A+B}{2}
\cot\frac{A-B}{2}
}
\]