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प्रश्न
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उत्तर
Given:
sin A + sin B = \[\frac{1}{4}\] .....(i)
cos A + cos B = \[\frac{1}{2}\] .....(ii)
Dividing (i) by (ii):
\[\Rightarrow \frac{\sin A + \sin B}{\cos A + \cos B} = \frac{\frac{1}{4}}{\frac{1}{2}}\]
\[ \Rightarrow \frac{2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)} = \frac{1}{2} \left[ \because \sin A + \sin B = 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)\text{ and }\cos A + \cos B = 2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right]\]
\[ \Rightarrow \frac{\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)} = \frac{1}{2}\]
\[ \Rightarrow \tan\left( \frac{A + B}{2} \right)=\frac{1}{2}\]
Hence proved.
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