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Transformation Formulae — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae5 Marks
Question
Prove that:
$\frac{\sin\text{A}+2\sin3\text{A}+\sin5\text{A}}{\sin3\text{A}+2\sin5\text{A}+\sin7\text{A}}=\frac{\sin3\text{A}}{\sin5\text{A}}$
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Answer

We have,

 $\text{LHS}=\frac{\sin\text{A}+2\sin3\text{A}+\sin5\text{A}}{\sin3\text{A}+2\sin5\text{A}+\sin7\text{A}}$

$=\ \frac{\sin5\text{A}+\sin\text{A}+2\sin3\text{A}}{\sin7\text{A}+\sin3\text{A}+2\sin5\text{A}}$

$=\ \frac{2\sin\Big(\frac{5\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-\text{A}}{2}\Big)+2\sin3\text{A}}{2\sin\Big(\frac{7\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-3\text{A}}{2}\Big)+2\sin5\text{A}}$

$=\ \frac{2\sin3\text{A}\cos2\text{A}+2\sin3\text{A}}{2\sin5\text{A}\cos2\text{A}+2\sin5\text{A}}$

$=\ \frac{2\sin3\text{A}(\cos2\text{A}+1)}{2\sin5\text{A}(\cos2\text{A}+1)}$

$=\ \frac{\sin3\text{A}}{\sin5\text{A}}$

$=\ \text{RHS}$

$\frac{\sin\text{A}+2\sin3\text{A}+\sin5\text{A}}{\sin3\text{A}+2\sin5\text{A}+\sin7\text{A}}=\frac{\sin3\text{A}}{\sin5\text{A}}$  Hence proved.

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