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Transformation Formulae — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae5 Marks
Question
Show that:

$\sin\text{A}\sin(\text{B}-\text{C})+\sin\text{B}\sin(\text{C}-\text{A})+\sin\text{C}\sin(\text{A}-\text{B})=0$

✓

Answer

We have,
$\text{LHS}=\ \sin\text{A}\sin(\text{B}-\text{C})+\sin\text{B}\sin(\text{C}-\text{A})+\sin\text{C}\sin(\text{A}-\text{B})$
$=\ \frac{1}{2}[2\sin\text{A}\sin(\text{B}-\text{C})+2\sin\text{B}\sin(\text{C}-\text{A})+2\sin\text{C}\sin(\text{A}-\text{B})]$
$=\ \frac{1}{2}\big[\cos(\text{A}-\text{B+C})-\cos(\text{A+B}-\text{C})+\cos(\text{B}-\text{C+A})\\\ \ \ \ \ \ \ \ -\cos(\text{B+C}-\text{A})+\cos(\text{C}-\text{A+B})-\cos(\text{C+A}-\text{B})\big]$
$=\ \frac{1}{2}\big[\cos(\text{A}-\text{B+C})-\cos(\text{A}-\text{B+C})-\cos(\text{A+B}-\text{C})\\\ \ \ \ \ \ \ \ +\cos(\text{A+B}-\text{C})-\cos(\text{B+C}-\text{A})+\cos(\text{B+C}-\text{A})\big]$
$=\ \frac{1}{2}\times0$
$=\ 0$
$=\ \text{RHS}$
$\therefore\ \sin\text{A}\sin(\text{B}-\text{C})+\sin\text{B}\sin(\text{C}-\text{A})+\sin\text{C}\sin(\text{A}-\text{B})$
$=0$
Hence proved.

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