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Transformation Formulae — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae4 Marks
Question
prove 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, $\text{LHS}=\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})$ $=\ \frac{1}{2}[2\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\cos(\text{B}-\text{D})+2\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})]$ $=\ \frac{1}{2}[\sin(\text{B}-\text{C}+\text{A}-\text{D})+\sin(\text{B}-\text{C}-\text{A+D})+\sin(\text{C}-\text{A}+\text{B}-\text{D})\\ \ \ \ \ +\sin(\text{C}-\text{A}-\text{B}+\text{D})+\sin(\text{A}-\text{B+C}-\text{D})+\sin(\text{A}-\text{B}-\text{C+D})]$ $=\ \frac{1}{2}[\sin(\text{A+B}-\text{C}-\text{D})+\sin(\text{B+D}-\text{C}-\text{A})+\sin(\text{B+C}-\text{A}-\text{D})\\ \ \ \ \ +\sin(\text{C+D}-\text{A}-\text{B})+\sin(\text{A+C}-\text{B}-\text{D})+\sin(\text{A+D}-\text{B}-\text{C})]$ $=\ \frac{1}{2}[\sin(\text{A+B}-\text{C}-\text{D})-\sin(\text{A+C}-\text{B}-\text{D})-\sin(\text{A+D}-\text{B}-\text{C})\\ \ \ \ \ -\sin(\text{A+B}-\text{C}-\text{D})+\sin(\text{A+C}-\text{B}-\text{D})+\sin(\text{A+D}-\text{B}-\text{C})]$ $=\ \frac{1}{2}[0]$ $=\ 0$ $=\ \text{RHS}$ $\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})=0$ Hence proved.

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