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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
$\text{a}\sin\frac{\text{A}}{2}\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)+\text{b}\sin\frac{\text{B}}{2}\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)+\text{c}\sin\frac{\text{C}}{2}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)=0.$
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Answer

$\text{LHS}=\text{a}\sin\frac{\text{A}}{2}\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)+\text{b}\sin\frac{\text{B}}{2}\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)\\+\text{c}\sin\frac{\text{C}}{2}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$
We know $\text{a}\sin\text{B = b}\sin\text{A,c}\sin\text{B = b}\sin\text{C},\\\text{a}\sin\text{C}=\text{c}\sin\text{B}$
$\text{a}\sin\frac{\text{A}}{2}\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)+\text{b}\sin\frac{\text{B}}{2}\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)\\+\text{c}\sin\frac{\text{C}}{2}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)=0$
$=\text{a}\sin\Big(\frac{\pi-(\text{B + C})}{2}\Big)\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)+\text{b}\sin\Big(\frac{\pi-(\text{C + A})}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)\\+\text{c}\sin\Big(\frac{\pi-(\text{A + B})}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$=\text{a}\cos\Big(\frac{\text{B + C}}{2}\Big)\sin\Big(\frac{\text{B}-\text{C}}{2}\Big)+\text{b}\cos\Big(\frac{\text{C + A}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)\\+\text{c}\cos\Big(\frac{\text{A + B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$=\text{a}(\sin\text{B}-\sin\text{C})+\text{b}(\sin\text{C}-\sin\text{A})+\text{c}(\sin\text{A}-\sin\text{B})$
$=\text{a}\sin\text{B}-\text{a}\sin\text{C}+\text{b}\sin\text{C}-\text{b}\sin\text{A + c}\sin\text{A}-\text{c}\sin\text{B}$
$=\text{b}\sin\text{A}-\text{a}\sin\text{C + b}\sin\text{C}-\text{b}\sin\text{A + a}\sin\text{C}-\text{b}\sin\text{C}$
$=0 =\text{RHS}$

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