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Question
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
Options
- \[- \frac{a}{b}\]
- \[- \frac{b}{a}\]
- \[\sqrt{a^2 + b^2}\]
none of these
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Solution
\[\text{ Given } : \]
\[sin\alpha + sin\beta = a\]
\[ \Rightarrow 2\sin\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2} = a . . . (1)\]
\[\text{ Also } , \]
\[cos\alpha + cos\beta = b\]
\[ \Rightarrow - 2\sin\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2} = b . . . (2)\]
\[\text{ On dividing (1) by (2), we get} \]
\[\frac{- \cos\frac{\alpha - \beta}{2}}{\sin\frac{\alpha - \beta}{2}} = \frac{a}{b}\]
\[ \Rightarrow \frac{- \sin\frac{\alpha - \beta}{2}}{\cos\frac{\alpha - \beta}{2}} = \frac{b}{a}\]
\[ \Rightarrow \tan\frac{\alpha - \beta}{2} = - \frac{b}{a}\]
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