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
(b) $-\frac{b}{a}$
Given:
sin α + sin β = a .....(i)
cos α − cos β = b .....(ii)
Dividing (i) by (ii):
$\Rightarrow \frac{\sin \alpha+\sin B}{\cos \alpha-\cos B}=\frac{a}{b}$
$\Rightarrow \frac{2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)}{-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)}=\frac{a}{b}$
$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right.$ and $\left.\cos A+\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]$
$\Rightarrow \frac{\sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)}{-\sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)}=\frac{a}{b}$
$\Rightarrow \cot \left(\frac{\alpha-\beta}{2}\right)=-\frac{a}{b}$
$\Rightarrow \frac{1}{\cot \left(\frac{\alpha-\beta}{2}\right)}=\frac{1}{-\frac{a}{b}}$
$\Rightarrow \tan \left(\frac{\alpha-\beta}{2}\right)=-\frac{b}{a}$