Prove that: + 3A+ 5A + 3A+ 5A = 3A - Vidyadip

Question

Prove that:
$\frac{\sin\text{A}+\sin3\text{A}+\sin5\text{A}}{\cos\text{A}+\cos3\text{A}+\cos5\text{A}}=\tan3\text{A}$

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Answer

We have,
$\text{LHS}=\frac{\sin\text{A}+\sin3\text{A}+\sin5\text{A}}{\cos\text{A}+\cos3\text{A}+\cos5\text{A}}$
$=\ \frac{(\sin5\text{A}+\sin\text{A})+\sin3\text{A}}{(\cos5\text{A}+\cos\text{A})+\cos3\text{A}}$
$=\ \frac{2\sin\Big(\frac{5\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-\text{A}}{2}\Big)+\sin3\text{A}}{2\cos\Big(\frac{5\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-\text{A}}{2}\Big)+\cos3\text{A}}$
$=\ \frac{2\sin3\text{A}\cos2\text{A}+\sin3\text{A}}{2\cos3\text{A}+\cos2\text{A}+\cos3\text{A}}$
$=\ \frac{\sin3\text{A}(2\cos2\text{A}+1)}{\cos3\text{A}(2\cos\text{A}+1)}$
$=\ \frac{\sin3\text{A}}{\cos3\text{A}}$
$=\ \tan3\text{A}$
$=\ \text{RHS}$
$\therefore\ \frac{\sin\text{A}+\sin3\text{A}+\sin5\text{A}}{\cos\text{A}+\cos3\text{A}+\cos5\text{A}}=\tan3\text{A}$ Hence proved.

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