- A$-\frac{\text{a}}{\text{b}}$
- B$-\frac{\text{b}}{\text{a}}$
- C$\sqrt{\text{a}^2+\text{b}^2}$
- DNone of these
Solution:
Given:
$\sin\alpha+\sin\beta=\text{a}$
$\Rightarrow2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}=\text{a}\ .....(1)$
Also,
$\cos\alpha+\cos\beta=\text{b}$
$\Rightarrow-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha+\beta}{2}=\text{b}\ .....(2)$
On dividing (1) by (2), we get
$\frac{-\cos\frac{\alpha+\beta}{2}}{\sin\frac{\alpha-\beta}{2}}=\frac{\text{a}}{\text{b}}$
$\Rightarrow\frac{-\sin\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}}=\frac{\text{b}}{\text{a}}$
$\Rightarrow\tan\frac{\alpha-\beta}{2}=-\frac{\text{b}}{\text{a}}$
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