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Transformation Formulae — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae4 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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