Question
$\frac{\text{a + b}}{\text{c}}=\frac{\cos\big(\frac{\text{A}-\text{B}}{2}\big)}{\sin\frac{\text{C}}{2}}$
✓
Answer
$\text{LHS}=\frac{\text{a + b}}{\text{c}}$
Let $\text{a = k}\sin\text{A, b = k}\sin\text{B, c = k}\sin\text{C}$
$=\frac{\text{k}\sin\text{A}+\text{k}\sin\text{B}}{\text{k}\sin\text{C}}$
$=\frac{\text{}\sin\text{A}+\text{}\sin\text{B}}{\text{}\sin\text{C}}$
$=\frac{2\sin\frac{\text{A + B}}{2}.\cos\frac{\text{A}-\text{B}}{2}}{2\sin\frac{\text{C}}{2}.\cos\frac{\text{C}}{2}}$
$=\frac{\sin\big(\frac{\pi-\text{C}}{2}\big).\cos\frac{\text{A}-\text{B}}{2}}{\sin\frac{\text{C}}{2}.\cos\frac{\text{C}}{2}}$
$=\frac{\cos\frac{\text{A}-\text{B}}{2}}{\sin\frac{\text{C}}{2}}=\text{RHS}$
Let $\text{a = k}\sin\text{A, b = k}\sin\text{B, c = k}\sin\text{C}$
$=\frac{\text{k}\sin\text{A}+\text{k}\sin\text{B}}{\text{k}\sin\text{C}}$
$=\frac{\text{}\sin\text{A}+\text{}\sin\text{B}}{\text{}\sin\text{C}}$
$=\frac{2\sin\frac{\text{A + B}}{2}.\cos\frac{\text{A}-\text{B}}{2}}{2\sin\frac{\text{C}}{2}.\cos\frac{\text{C}}{2}}$
$=\frac{\sin\big(\frac{\pi-\text{C}}{2}\big).\cos\frac{\text{A}-\text{B}}{2}}{\sin\frac{\text{C}}{2}.\cos\frac{\text{C}}{2}}$
$=\frac{\cos\frac{\text{A}-\text{B}}{2}}{\sin\frac{\text{C}}{2}}=\text{RHS}$
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