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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae4 Marks
Question
Prove that:
$\frac{\sin5\text{A}-\sin7\text{A}+\sin8\text{A}-\sin4\text{A}}{\cos4\text{A}+\cos7\text{A}-\cos5\text{A}-\cos8\text{A}}=\cot6\text{A}$
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Answer

We have,
$\text{LHS}=\frac{\sin5\text{A}-\sin7\text{A}+\sin8\text{A}-\sin4\text{A}}{\cos4\text{A}+\cos7\text{A}-\cos5\text{A}-\cos8\text{A}}$
$=\ \frac{-(\sin7\text{A}-\sin5\text{A})+(\sin8\text{A}-\sin4\text{A})}{-(\cos7\text{A}-\cos5\text{A})-(\cos8\text{A}-\cos4\text{A})}$
$=\ \frac{-\Big[2\sin\Big(\frac{7\text{A}-5\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\Big]+\Big[2\sin\Big(\frac{8\text{A}-4\text{A}}{2}\Big)\cos\Big(\frac{8\text{A}+4\text{A}}{2}\Big)\Big]}{-2\sin\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\sin\Big(\frac{7\text{A}-5\text{A}}{2}\Big)-\Big[-2\sin\Big(\frac{8\text{A}+4\text{A}}{2}\Big)\sin\Big(\frac{8\text{A}-4\text{A}}{2}\Big)\Big]}$
$=\ \frac{-2\sin\text{A}\cos6\text{A}+2\sin2\text{A}\cos6\text{A}}{-2\sin6\text{A}\sin\text{A}+2\sin6\text{A}\sin2\text{A}}$
$=\ \frac{2\cos6\text{A}[-\sin\text{A}+\sin2\text{A}]}{-2\sin6\text{A}[-\sin\text{A}+\sin2\text{A}]}$
$=\ \frac{\cos6\text{A}}{\sin6\text{A}}$
$=\ \cot6\text{A}$
$=\ \text{RHS}$
$\frac{\sin5\text{A}-\sin7\text{A}+\sin8\text{A}-\sin4\text{A}}{\cos4\text{A}+\cos7\text{A}-\cos5\text{A}-\cos8\text{A}}=\cot6\text{A}$ Hence proved.

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