Question

If sin α + sin β = a and cos α − cos β = b, then tan $\frac{\mathrm{\alpha }-\mathrm{\beta }}{2}$=
(a) $-\frac{a}{b}$

(b) $-\frac{b}{a}$

(c) $\sqrt{a^2+b^2}$

(d) None of these

Answer 1

(b) $-\frac{b}{a}$

Given:
sin α + sin β = a .....(i)
cos α − cos β = b .....(ii)

Dividing (i) by (ii):

$$\begin{gathered} \Rightarrow \frac{\sin \alpha +\sin B}{\cos \alpha -\cos B}=\frac{a}{b} \\ \Rightarrow \frac{2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)}{-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)}=\frac{a}{b}\left[\because \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\mathrm{and}\cos A+\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)\right] \\ \Rightarrow \frac{\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)}{-\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)}=\frac{a}{b} \\ \Rightarrow \cot \left(\frac{\alpha -\beta }{2}\right)=-\frac{a}{b} \\ \Rightarrow \frac{1}{\cot \left(\frac{\alpha -\beta }{2}\right)}=\frac{1}{-{\displaystyle \frac{a}{b}}} \\ \Rightarrow \tan \left({\displaystyle \frac{\alpha -\beta }{2}}\right)=-\frac{b}{a} \end{gathered}$$