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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}-\cos\text{A}}=\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$
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Answer

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

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