If \( \sin\alpha+\sin\beta=a \) and \( \cos\alpha-\cos\beta=b \), Find \( \tan\frac{\alpha-\beta}{2} \)
Question
If
\[ \sin\alpha+\sin\beta=a \]
and
\[ \cos\alpha-\cos\beta=b, \]
then find
\[ \tan\frac{\alpha-\beta}{2} \]
(a) \(-\frac{a}{b}\)
(b) \(-\frac{b}{a}\)
(c) \(\sqrt{a^2+b^2}\)
(d) none of these
Solution
Using the sum-to-product identities,
\[ \sin\alpha+\sin\beta = 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \]
Therefore,
\[ a = 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \]
Also,
\[ \cos\alpha-\cos\beta = -2\sin\frac{\alpha+\beta}{2} \sin\frac{\alpha-\beta}{2} \]
Hence,
\[ b = -2\sin\frac{\alpha+\beta}{2} \sin\frac{\alpha-\beta}{2} \]
Dividing the second equation by the first,
\[ \frac{b}{a} = \frac{-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}} {2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}} \]
\[ = -\tan\frac{\alpha-\beta}{2} \]
Therefore,
\[ \tan\frac{\alpha-\beta}{2} = -\frac{b}{a} \]
Final Answer
\[ \boxed{\tan\frac{\alpha-\beta}{2}=-\frac{b}{a}} \]
Hence, the correct option is (b) \(-\frac{b}{a}\).