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Question
If sin α + sin β = a and cos α − cos β = b, 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
Given:
sin α + sin β = a .....(i)
cos α − cos β = b .....(ii)
Dividing (i) by (ii):
\[\Rightarrow \frac{\sin\alpha + \sin B}{\cos\alpha - \cos B} = \frac{a}{b}\]
\[ \Rightarrow \frac{2\sin\left( \frac{\alpha + \beta}{2} \right)\cos\left( \frac{\alpha - \beta}{2} \right)}{- 2\sin\left( \frac{\alpha + \beta}{2} \right)\sin\left( \frac{\alpha - \beta}{2} \right)} = \frac{a}{b} \left[ \because \sin A + \sin B = 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \text{ and }\cos A + \cos B = - 2\sin\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right) \right]\]
\[ \Rightarrow \frac{\sin\left( \frac{\alpha + \beta}{2} \right)\cos\left( \frac{\alpha - \beta}{2} \right)}{- \sin\left( \frac{\alpha + \beta}{2} \right)\sin\left( \frac{\alpha - \beta}{2} \right)} = \frac{a}{b}\]
\[ \Rightarrow \cot\left( \frac{\alpha - \beta}{2} \right)=-\frac{a}{b}\]
\[ \Rightarrow \frac{1}{\cot\left( \frac{\alpha - \beta}{2} \right)}=\frac{1}{- \frac{a}{b}}\]
\[ \Rightarrow \tan\left( \frac{\alpha - \beta}{2} \right)=-\frac{b}{a}\]
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