Answer: see proof below
Step-by-step explanation:
Use the following Product to Sum Identities:
2 sin A · cos B = sin (A + B) + sin (A - B)
2 cos A · cos B = sin (A + B) + sin (A - B)
Given: cos A + cos B = 1/2 and sin A + sin B = 1/4
Proof LHS → RHS
[tex]\text{LHS:}\qquad \qquad \qquad \tan\dfrac{A+B}{2}[/tex]
[tex]\text{Expand:}\qquad \qquad \dfrac{\sin\frac{(A+B)}{2}}{\cos\frac{(A+B)}{2}}[/tex]
[tex]\text{Multiplication:}\qquad \quad \dfrac{\sin\frac{(A+B)}{2}}{\cos\frac{(A+B)}{2}}\bigg(\dfrac{2\cos\frac{A-B}{2}}{2\cos \frac{A-B}{2}}\bigg)[/tex]
[tex]\text{Simplify:}\qquad \qquad \quad \dfrac{2\sin \frac{A+B}{2}\cdot \cos \frac{A-B}{2}}{2\cos \frac{A+B}{2}\cdot \cos \frac{A-B}{2}}[/tex]
[tex]\text{Product to Sum:}\qquad \dfrac{\sin A+\sin B}{\cos A+\cos B}[/tex]
[tex]\text{Given:}\qquad \qquad \qquad \quad \dfrac{\frac{1}{4}}{\frac{1}{2}}[/tex]
[tex]\text{Simplify:}\qquad \qquad \qquad \dfrac{1}{2}[/tex]
LHS = RHS: 1/2 = 1/2 [tex]\checkmark[/tex]
