If sin α + sin β = a and cos α − cos β = b,

(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 $\left.\cos A+\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\right]$

$\Rightarrow \frac{\sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)}{-\sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)}=\frac{a}{b}$

$\Rightarrow \cot \left(\frac{\alpha-\beta}{2}\right)=-\frac{a}{b}$

$\Rightarrow \frac{1}{\cot \left(\frac{\alpha-\beta}{2}\right)}=\frac{1}{-\frac{a}{b}}$

$\Rightarrow \tan \left(\frac{\alpha-\beta}{2}\right)=-\frac{b}{a}$