Prove that: 4A+ 3A+ 2A 4A+ 3A+ 2A = 3A - Vidyadip

Question

Prove that: $\frac{\cos4\text{A}+\cos3\text{A}+\cos2\text{A}}{\sin4\text{A}+\sin3\text{A}+\sin2\text{A}}=\cot3\text{A}$

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Answer

We have, $\text{LHS}=\frac{\cos4\text{A}+\cos3\text{A}+\cos2\text{A}}{\sin4\text{A}+\sin2\text{A}+\sin2\text{A}}$ $=\ \frac{(\cos4\text{A}+\cos2\text{A})+\cos3\text{A}}{(\sin4\text{A}+\sin2\text{A})+\sin3\text{A}}$ $=\ \frac{2\cos\Big(\frac{4\text{A}+2\text{A}}{2}\Big)\cos\Big(\frac{4\text{A}-2\text{A}}{2}\Big)+\cos3\text{A}}{2\sin\Big(\frac{4\text{A}+2\text{A}}{2}\Big)\cos\Big(\frac{4\text{A}-2\text{A}}{2}\Big)+\sin3\text{A}}$ $=\ \frac{2\cos3\text{A}\cos\text{A}+\cos3\text{A}}{2\sin3\text{A}\cos\text{A}+\sin3\text{A}}$ $=\ \frac{\cos3\text{A}(2\cos\text{A}+1)}{\sin3\text{A}(2\cos2\text{A}+1)}$ $=\ \frac{\cos3\text{A}}{\sin\text{A}}$ $=\ \cot3\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\cos4\text{A}+\cos3\text{A}+\cos2\text{A}}{\sin4\text{A}+\sin3\text{A}+\sin2\text{A}}=\cot3\text{A}$ Hence proved.

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