Question

Someone please help me to prove this. ​

Answer 1

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

$\text{LHS:}\qquad \qquad \qquad \tan\dfrac{A+B}{2}$

$\text{Expand:}\qquad \qquad \dfrac{\sin\frac{(A+B)}{2}}{\cos\frac{(A+B)}{2}}$

$\text{Multiplication:}\qquad \quad \dfrac{\sin\frac{(A+B)}{2}}{\cos\frac{(A+B)}{2}}\bigg(\dfrac{2\cos\frac{A-B}{2}}{2\cos \frac{A-B}{2}}\bigg)$

$\text{Simplify:}\qquad \qquad \quad \dfrac{2\sin \frac{A+B}{2}\cdot \cos \frac{A-B}{2}}{2\cos \frac{A+B}{2}\cdot \cos \frac{A-B}{2}}$

$\text{Product to Sum:}\qquad \dfrac{\sin A+\sin B}{\cos A+\cos B}$

$\text{Given:}\qquad \qquad \qquad \quad \dfrac{\frac{1}{4}}{\frac{1}{2}}$

$\text{Simplify:}\qquad \qquad \qquad \dfrac{1}{2}$

LHS = RHS:    1/2 = 1/2  $\checkmark$