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Transformation Formulae — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae5 Marks
Question
Prove that:
$\frac{\sin\text{A}+\sin\text{B}}{\sin\text{A}-\sin\text{B}}=\tan\Big(\frac{\text{A}-\text{B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$
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Answer

We have,
$\text{LHS}=\frac{\sin\text{A}+\sin\text{B}}{\sin\text{A}-\sin\text{B}}$
$=\ \frac{2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{2\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)}$
$=\ \frac{\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)}$
$= \tan\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$
$=\ \text{RHS}$
$\therefore\ \frac{\sin\text{A}+\sin\text{B}}{\sin\text{A}-\sin\text{B}}=\tan\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big).$ Hence proved.

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