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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
Prove that:
$\frac{\sin5\text{A}\cos2\text{A}-\sin6\text{A}\cos\text{A}}{\sin\text{A}\sin2\text{A}-\cos2\text{A}\cos3\text{A}}=\tan\text{A}$
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Answer

We have,
$\text{LHS}=\frac{\sin5\text{A}\cos2\text{A}-\sin6\text{A}\cos\text{A}}{\sin\text{A}\sin2\text{A}-\cos2\text{A}\cos3\text{A}}$
$=\ \frac{2(\sin5\text{A}\cos2\text{A}-\sin6\text{A}\cos\text{A})}{2(\sin\text{A}\sin2\text{A}-\cos2\text{A}\cos3\text{A})}$
$=\ \frac{2\sin5\text{A}\cos2\text{A}-2\sin6\text{A}\cos\text{A}}{2\sin\text{A}\sin2\text{A}-2\cos2\text{A}\cos3\text{A}}$
$=\ \frac{\sin(5\text{A}+2\text{A})+\sin(5\text{A}-2\text{A})-[\sin(6\text{A}+\text{A})+\sin(6\text{A}-\text{A})]}{\cos(2\text{A}-\text{A})-\cos(2\text{A}+\text{A})-[\cos(3\text{A}+2\text{A})+\cos(3\text{A}-2\text{A})]}$
$=\ \frac{\sin7\text{A}+\sin3\text{A}-\sin7\text{A}-\sin5\text{A}}{\cos\text{A}-\cos3\text{A}-\cos5\text{A}-\cos\text{A}}$
$=\ \frac{\sin3\text{A}-\sin5\text{A}}{-\cos3\text{A}-\cos5\text{A}}$
$=\ \frac{-(\sin5\text{A}-\sin3\text{A})}{-(\cos5\text{A}+\cos3\text{A})}$
$=\ \frac{\sin5\text{A}-\sin3\text{A}}{\cos5\text{A}+\cos3\text{A}}$
$=\ \frac{2\sin\Big(\frac{5\text{A}-3\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}+3\text{A}}{2}\Big)}{2\cos\Big(\frac{5\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-3\text{A}}{2}\Big)}$
$=\ \frac{\sin\text{A}\cos4\text{A}}{\cos4\text{A}\cos\text{A}}$
$=\ \frac{\sin\text{A}}{\cos\text{A}}$
$=\ \tan\text{A}$
$=\ \text{RHS}$
$\therefore\ \frac{\sin5\text{A}\cos2\text{A}-\sin6\text{A}\cos\text{A}}{\sin\text{A}\sin2\text{A}-\cos2\text{A}\cos3\text{A}}=\tan\text{A}$ Hence proved.

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