Question
Prove that: $\frac{\cos3\text{A}+2\cos5\text{A}+\cos7\text{A}}{\cos\text{A}+2\cos3\text{A}+\cos5\text{A}}=\frac{\cos5\text{A}}{\cos3\text{A}}$

Answer

We have, $\text{LHS}=\frac{\cos3\text{A}+2\cos5\text{A}+\cos7\text{A}}{2\cos\text{A}+2\cos3\text{A}+\cos5\text{A}}$ $=\ \frac{(\cos7\text{A}+\cos3\text{A})+2\cos5\text{A}}{(\cos5\text{A}+\cos\text{A})+2\cos3\text{A}}$ $=\ \frac{2\cos\Big(\frac{7\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-3\text{A}}{2}\Big)+2\cos5\text{A}}{2\cos\Big(\frac{5\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-\text{A}}{2}\Big)+\cos3\text{A}}$ $=\ \frac{2\cos5\text{A}\cos2\text{A}+2\cos5\text{A}}{2\cos3\text{A}\cos2\text{A}+2\cos3\text{A}}$ $=\ \frac{2\cos5\text{A}(\cos2\text{A}+1)}{2\cos3\text{A}(\cos2\text{A}+1)}$ $=\ \frac{\cos5\text{A}}{\cos3\text{A}}$ $=\ \tan3\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\cos3\text{A}+2\sin5\text{A}+\cos7\text{A}}{\cos\text{A}+2\cos3\text{A}+\cos5\text{A}}=\frac{\cos5\text{A}}{\cos3\text{A}}$ Hence proved.

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