Question:
If $\sin \alpha+\sin \beta=a$ and $\cos \alpha-\cos \beta=b$ then $\tan \frac{\alpha-\beta}{2}=$
(a) $-\frac{a}{b}$
(b) $-\frac{b}{a}$
(c) $\sqrt{a^{2}+b^{2}}$
(d) none of these
Solution:
(b) $-\frac{b}{a}$
Given:
$\sin \alpha+\sin \beta=a$
$\Rightarrow 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=a \quad \ldots(1)$
Also,
$\cos \alpha+\cos \beta=b$
$\Rightarrow-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}=b \quad \ldots(2)$
On dividing $(1)$ by $(2)$, we get
$\frac{-\cos \frac{\alpha-\beta}{2}}{\sin \frac{\alpha-\beta}{2}}=\frac{a}{b}$
$\Rightarrow \frac{-\sin \frac{\alpha-\beta}{2}}{\cos \frac{\alpha-\beta}{2}}=\frac{b}{a}$
$\Rightarrow \tan \frac{\alpha-\beta}{2}=-\frac{b}{a}$