Prove that: (A-B) + (B-C) + (C-A) =0

Question

Prove that: $\frac{\sin\text{(A-B)}}{\sin\text{A}\sin\text{B}}+\frac{\sin\text{(B-C)}}{\sin\text{B}\sin\text{C}}+\frac{\sin\text{(C-A)}}{\sin\text{C}\sin\text{A}}=0$

✓

Answer

We have,
$\text{L.H.S }\frac{\sin\text{(A-B)}}{\sin\text{A}\sin\text{B}}+\frac{\sin\text{(B-C)}}{\sin\text{B}\sin\text{C}}+\frac{\sin\text{(C-A)}}{\sin\text{C}\sin\text{A}}$
$=\frac{\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}}{\sin\text{A}\sin\text{B}}+\frac{\sin\text{B}\cos\text{C}-\cos\text{B}\sin\text{C}}{\sin\text{B}\sin\text{C}}\\\ \ +\frac{\sin\text{C}\cos\text{A}-\cos\text{C}\sin\text{A}}{\sin\text{C}\sin\text{A}}$
$=\frac{\sin\text{A}\cos\text{B}}{\sin\text{A}\sin\text{B}}-\frac{\cos\text{A}\sin\text{A}}{\sin\text{A}\sin\text{B}}+\frac{\sin\text{B}\cos\text{C}}{\sin\text{B}\sin\text{C}}\\\ \ -\frac{\cos\text{B}\sin\text{C}}{\sin\text{B}\sin\text{C}}+\frac{\sin\text{C}\cos\text{A}}{\sin\text{C}\sin\text{A}}-\frac{\cos\text{C}\sin\text{A}}{\sin\text{C}\sin\text{A}}$
$=\frac{\cos\text{B}}{\sin\text{B}}-\frac{\cos\text{A}}{\sin\text{A}}+\frac{\cos\text{C}}{\sin\text{C}}-\frac{\cos\text{B}}{\sin\text{B}}+\frac{\cos\text{A}}{\sin\text{A}}-\frac{\cos\text{C}}{\sin\text{C}}$ $\Big[\because\cot\theta=\frac{\cos\theta}{\sin\theta}\Big]$
$=\cot\text{B}-\cot\text{A}+\cot\text{C}-\cot\text{B}+\cot\text{A}-\cot\text{C}$
$=0$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.