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Transformation Formulae — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae5 Marks
Question
Show that:

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

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Answer

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

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