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Trigonometric Ratios Of Multiple And Sub Multiple Angles — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Multiple And Sub Multiple Angles5 Marks
Question
Prove that: $\frac{\sin\text{(A-B)}}{\sin\text{A}\sin\text{B}}+\frac{\sin\text{(B-C)}}{\sin\text{B}\sin\text{C}}+\frac{\sin\text{(C-A)}}{\sin\text{C}\sin\text{A}}=0$
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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.

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