Question

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

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

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

(d) none of these

Answer 1

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

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{sin\alpha }+\mathrm{sin\beta }=a \\ \Rightarrow 2\sin \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\cos \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}=a...\left(1\right) \\ \mathrm{Also}, \\ \mathrm{cos\alpha }+\mathrm{cos\beta }=b \\ \Rightarrow -2\sin \frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\sin \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}=b...\left(2\right) \\ \mathrm{On}\mathrm{dividing}\left(1\right)\mathrm{by}\left(2\right),\mathrm{we}\mathrm{get} \\ \frac{-\cos {\displaystyle \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}}}{\sin {\displaystyle \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}}}=\frac{a}{b} \\ \Rightarrow \frac{-\sin {\displaystyle \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}}}{\cos {\displaystyle \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}}}=\frac{b}{a} \\ \Rightarrow \tan \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}=-\frac{b}{a} \end{gathered}$$