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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\frac{\sin3\text{A}+\sin5\text{A}+\sin7\text{A}+\sin9\text{A}}{\cos3\text{A}+\cos5\text{A}+\cos7\text{A}\cos9\text{A}}=\tan6\text{A}$
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Answer

We have,
$\text{LHS}=\frac{\sin3\text{A}+\sin5\text{A}+\sin7\text{A}+\sin9\text{A}}{\cos3\text{A}+\cos5\text{A}+\cos7\text{A}+\cos9\text{A}}$
$=\ \frac{(\sin9\text{A}+\sin3\text{A})+(\sin7\text{A}+\sin5\text{A})}{(\cos9\text{A}+\cos3\text{A})+(\cos7\text{A}+\cos5\text{A})}$
$=\ \frac{2\sin\Big(\frac{9\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{9\text{A}-3\text{A}}{2}\Big)+2\sin\Big(\frac{9\text{A}-3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-5\text{A}}{2}\Big)}{2\cos\Big(\frac{9\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{9\text{A}-3\text{A}}{2}\Big)+2\cos\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-5\text{A}}{2}\Big)}$
$=\ \frac{2\sin6\text{A}\cos3\text{A}+2\sin6\text{A}\cos\text{A}}{2\cos6\text{A}\cos3\text{A}+2\cos6\text{A}\cos\text{A}}$
$=\ \frac{2\sin6\text{A}(\cos3\text{A}+\cos\text{A})}{2\cos6\text{A}(\cos3\text{A}+\cos\text{A})}$
$=\ \frac{\sin6\text{A}}{\cos6\text{A}}$
$=\ \tan6\text{A}$
$=\ \text{RHS}$
$\therefore\ \frac{\sin3\text{A}+\sin5\text{A}+\sin7\text{A}+\sin9\text{A}}{\cos3\text{A}+\cos5\text{A}+\cos7\text{A}\cos9\text{A}}=\tan6\text{A}$ Hence proved.

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