If \( \sin A+\sin B=\alpha \) and \( \cos A+\cos B=\beta \), then write the value of \( \tan\frac{A+B}{2} \)
Solution:
Using identities,
\[
\sin A+\sin B
=
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
\]
\[
\alpha
=
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
\]
Also,
\[
\cos A+\cos B
=
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\]
\[
\beta
=
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\]
Dividing,
\[
\frac{\alpha}{\beta}
=
\frac{
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
}{
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
}
\]
\[
\frac{\alpha}{\beta}
=
\tan\frac{A+B}{2}
\]
Therefore,
\[
\boxed{\tan\frac{A+B}{2}=\frac{\alpha}{\beta}}
\]