Prove that: 3A 4A- 2A 4A + 6A = 2A - Vidyadip

Question

Prove that: $\frac{\sin3\text{A}\cos4\text{A}-\sin\text{A}\cos2\text{A}}{\sin4\text{A}\sin\text{A}+\cos6\text{A}\cos\text{A}}=\tan2\text{A}$

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Answer

We have, $\text{LHS}=\frac{\sin3\text{A}\cos4\text{A}-\sin\text{A}\cos2\text{A}}{\sin4\text{A}\sin\text{A}+\cos6\text{A}\cos\text{A}}$ $=\ \frac{2(\sin3\text{A}\cos4\text{A}-\sin\text{A}\cos2\text{A})}{2(\sin4\text{A}\sin\text{A}+\cos6\text{A}\cos\text{A})}$ $=\ \frac{2\sin3\text{A}\cos4\text{A}-2\sin\text{A}\cos2\text{A}}{2\sin4\text{A}\sin\text{A}+2\cos6\text{A}\cos\text{A}}$ $=\ \frac{\sin(4\text{A}+3\text{A})-\sin(4\text{A}-3\text{A})-[\sin(2\text{A}+\text{A})-\sin(2\text{A}-\text{A})]}{\cos(4\text{A}-\text{A})-\cos(4\text{A}+\text{A})+\cos(6\text{A}+\text{A})+\cos(6\text{A}-\text{A})}$ $=\ \frac{\sin(7\text{A})-\sin(\text{A})-\sin(3\text{A})+\sin(\text{A})}{\cos(3\text{A})-\cos(5\text{A})+\cos(7\text{A})+\cos(5\text{A})}$ $=\ \frac{\sin(7\text{A})-\sin(3\text{A})}{\cos(3\text{A})+\cos(7\text{A})}$ $=\ \frac{2\sin\Big(\frac{7\text{A}-3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}+3\text{A}}{2}\Big)}{2\cos\Big(\frac{7\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-3\text{A}}{2}\Big)}$ $=\ \frac{\sin2\text{A}}{\cos2\text{A}}$ $=\ \tan2\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\sin3\text{A}\cos4\text{A}-\sin\text{A}\cos2\text{A}}{\sin4\text{A}\sin\text{A}+\cos6\text{A}\cos\text{A}}=\tan2\text{A}$ Hence proved.

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