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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\frac{\sin11\text{A}\sin\text{A}+\sin7\text{A}\sin3\text{A}}{\cos11\text{A}\sin\text{A}+\cos7\text{A}\sin3\text{A}}=\tan8\text{A}$
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Answer

We have,
$\text{LHS}=\frac{\sin11\text{A}\sin\text{A}+\sin7\text{A}\sin3\text{A}}{\cos11\text{A}\sin\text{A}+\cos7\text{A}\sin3\text{A}}$
$=\ \frac{2(\sin11\text{A}\sin\text{A}+\sin7\text{A}\sin3\text{A})}{2(\cos11\text{A}\sin\text{A}+\cos7\text{A}\sin3\text{A})}$
$=\ \frac{2\sin11\text{A}\sin\text{A}+2\sin7\text{A}\sin3\text{A}}{2\sin11\text{A}\sin\text{A}+2\cos7\text{A}\sin3\text{A}}$
$=\ \frac{\cos(11\text{A}-\text{A})-\cos(11\text{A}+\text{A})+\cos(7\text{A}-3\text{A})-\cos(7\text{A}+3\text{A})}{\sin(11\text{A}+\text{A})-\sin(11\text{A}-\text{A})+\sin(7\text{A}+3\text{A})-\sin(7\text{A}-3\text{A})]}$
$=\ \frac{\cos10\text{A}-\cos12\text{A}+\cos4\text{A}-\cos10\text{A}}{\sin12\text{A}-\sin10\text{A}+\sin10\text{A}-\sin4\text{A}}$
$=\ \frac{-(\cos12\text{A}-\cos4\text{A})}{\sin12\text{A}-\sin4\text{A}}$
$=\ \frac{-\Big[2\sin\Big(\frac{12\text{A}+4\text{A}}{2}\Big)\sin\Big(\frac{12\text{A}-4\text{A}}{2}\Big)\Big]}{2\sin\Big(\frac{12\text{A}-4\text{A}}{2}\Big)\cos\Big(\frac{12\text{A}+4\text{A}}{2}\Big)}$
$=\ \frac{2\sin8\text{A}\sin4\text{A}}{2\sin4\text{A}\cos8\text{A}}$
$=\ \frac{\sin8\text{A}}{\cos8\text{A}}$
$=\ \tan8\text{A}$
$=\ \text{RHS}$
$\therefore\ \frac{\sin11\text{A}\sin\text{A}+\sin7\text{A}\sin3\text{A}}{\cos11\text{A}\sin\text{A}+\cos7\text{A}\sin3\text{A}}=\tan8\text{A}$ Hence proved.

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