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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\frac{\sin\text{A}+\sin\text3{A}}{\cos\text{A}-\cos3\text{A}}=\cot\text{A}$
✓

Answer

We have,
$\text{LHS}=\frac{\sin\text{A}+\sin3\text{A}}{\cos\text{A}-\cos3\text{A}}$
$=\ \frac{2\sin\Big(\frac{\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{\text{A}-3\text{A}}{2}}{-2\sin\Big(\frac{\text{A}+3\text{A}}{2}\Big)\sin\Big(\frac{\text{A}-3\text{A}}{2}\Big)}$
$=\ \frac{-\sin2\text{A}\times\cos(-\text{A})}{\sin2\text{A}\sin(-\text{A})}$
$=\ \frac{-\cos(-\text{A})}{\sin(-\text{A})}$
$=\ \frac{-\cos\text{A}}{-\sin\text{A}}$ $[\because\ \cos(-\theta)=\cos\theta\text{ and }\sin(-\theta)=-\sin\theta]$
$=\ \frac{\cos\text{A}}{\sin\text{A}}$
$=\ \cot\text{A}$
$=\ \text{RHS}$
$\therefore\ \frac{\sin\text{A}+\sin3\text{A}}{\cos\text{A}-\cos3\text{A}}=\cot\text{A}.$ Hence proved.

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