Question

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

Answer 1

Given that cos A + cos B = ${\displaystyle \frac{1}{2}}$

2 cos ${\displaystyle \left(\frac{\text{A + B}}{2}\right)\cos \left(\frac{\text{A - B}}{2}\right)=\frac{1}{2}}$   ..(1)

Also given that sin A + sin B = ${\displaystyle \frac{1}{4}}$

2 sin ${\displaystyle \left(\frac{\text{A + B}}{2}\right)\cos \left(\frac{\text{A - B}}{2}\right)=\frac{1}{4}}$   ..(2)

${\displaystyle \left(2\right)÷\left(1\right)}$ we get,

${\displaystyle \frac{2\sin \left(\frac{\text{A + B}}{2}\right)\cos \left(\frac{\text{A - B}}{2}\right)}{2\cos \left(\frac{\text{A + B}}{2}\right)\cos \left(\frac{\text{A - B}}{2}\right)}=\frac{\frac{1}{4}}{\frac{1}{2}}}$

tan ${\displaystyle \left(\frac{\text{A + B}}{2}\right)}$ = ${\displaystyle \frac{2}{4}}$

∴ tan ${\displaystyle \left(\frac{\text{A + B}}{2}\right)=\frac{1}{2}}$