If \( \sin\alpha+\sin\beta=a \) and \( \cos\alpha-\cos\beta=b \), then \( \tan\frac{\alpha-\beta}{2} \) is
Options:
(a) \( -\frac{a}{b} \)
(b) \( -\frac{b}{a} \)
(c) \( \sqrt{a^2+b^2} \)
(d) none of these
Solution:
Using identities,
\[
\sin C+\sin D
=
2\sin\frac{C+D}{2}\cos\frac{C-D}{2}
\]
\[
a
=
2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}
\]
Also,
\[
\cos C-\cos D
=
-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}
\]
\[
b
=
-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}
\]
Dividing,
\[
\frac{b}{a}
=
\frac{-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}}
{2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}}
\]
\[
\frac{b}{a}
=
-\tan\frac{\alpha-\beta}{2}
\]
\[
\tan\frac{\alpha-\beta}{2}
=
-\frac{b}{a}
\]
\[
\boxed{-\frac{b}{a}}
\]
Correct option: (b)