Question

\[\text{ If } \cos A + \cos B = \frac{1}{2}\text{ and }\sin A + \sin B = \frac{1}{4},\text{ prove that }\tan\left( \frac{A + B}{2} \right) = \frac{1}{2} .\]

 

Answer 1

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.