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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\frac{\cos3\text{A}+2\cos5\text{A}+\cos7\text{A}}{\cos\text{A}+2\cos3\text{A}+\cos5\text{A}}=\frac{\cos5\text{A}}{\cos3\text{A}}$
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Answer

We have,
$\text{LHS}=\frac{\cos3\text{A}+2\cos5\text{A}+\cos7\text{A}}{2\cos\text{A}+2\cos3\text{A}+\cos5\text{A}}$
$=\ \frac{(\cos7\text{A}+\cos3\text{A})+2\cos5\text{A}}{(\cos5\text{A}+\cos\text{A})+2\cos3\text{A}}$
$=\ \frac{2\cos\Big(\frac{7\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-3\text{A}}{2}\Big)+2\cos5\text{A}}{2\cos\Big(\frac{5\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-\text{A}}{2}\Big)+\cos3\text{A}}$
$=\ \frac{2\cos5\text{A}\cos2\text{A}+2\cos5\text{A}}{2\cos3\text{A}\cos2\text{A}+2\cos3\text{A}}$
$=\ \frac{2\cos5\text{A}(\cos2\text{A}+1)}{2\cos3\text{A}(\cos2\text{A}+1)}$
$=\ \frac{\cos5\text{A}}{\cos3\text{A}}$
$=\ \tan3\text{A}$
$=\ \text{RHS}$
$\therefore\ \frac{\cos3\text{A}+2\sin5\text{A}+\cos7\text{A}}{\cos\text{A}+2\cos3\text{A}+\cos5\text{A}}=\frac{\cos5\text{A}}{\cos3\text{A}}$ Hence proved.

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