Prove that: 2A+ 3A 6A 2A+ 3A 6A = 5A - Vidyadip

Question

Prove that: $\frac{\sin\text{A}\sin2\text{A}+\sin3\text{A}\sin6\text{A}}{\sin\text{A}\cos2\text{A}+\sin3\text{A}\cos6\text{A}}=\tan5\text{A}$

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Answer

We have, $\text{LHS}=\frac{\sin\text{A}\sin2\text{A}+\sin3\text{A}\sin6\text{A}}{\sin\text{A}\cos2\text{A}+\sin3\text{A}\cos6\text{A}}$ $=\ \frac{2[\sin\text{A}\sin2\text{A}+\sin3\text{A}\sin6\text{A}]}{2[\sin\text{A}\cos2\text{A}+\sin3\text{A}\cos6\text{A}]}$ $=\ \frac{2\sin2\text{A}\sin\text{A}+2\sin6\text{A}\sin3\text{A}}{2\cos2\text{A}\sin\text{A}+2\cos6\text{A}\sin3\text{A}}$ $=\ \frac{\cos(2\text{A}-\text{A})-\cos(2\text{A}+\text{A})+\cos(6\text{A}-3\text{A})-\cos(6\text{A}+3\text{A})}{\sin(2\text{A}+\text{A})-\sin(2\text{A}-\text{A})+\sin(6\text{A}+3\text{A})+\sin(6\text{A}-3\text{A})}$ $=\ \frac{\cos\text{A}-\cos3\text{A}+\cos3\text{A}-\cos9\text{A}}{\sin3\text{A}-\sin\text{A}+\sin9\text{A}-\sin3\text{A}}$ $=\ \frac{\cos\text{A}-\cos9\text{A}}{\sin9\text{A}-\sin\text{A}}$ $=\ \frac{-[\cos9\text{A}-\cos\text{A}]}{\sin9\text{A}-\sin\text{A}}$ $=\ \frac{-\Big(-2\sin\Big(\frac{9\text{A}+\text{A}}{2}\Big)\times\sin\Big(\frac{9\text{A}-\text{A}}{2}\Big)\Big)}{2\sin\Big(\frac{9\text{A}-\text{A}}{2}\Big)\times\cos\Big(\frac{9\text{A}+\text{A}}{2}\Big)}$ $=\ \frac{\sin5\text{A}\sin4\text{A}}{\sin4\text{A}\cos5\text{A}}$ $=\ \tan5\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\sin\text{A}\sin2\text{A}+\sin3\text{A}\sin6\text{A}}{\sin\text{A}\cos2\text{A}+\sin3\text{A}\cos6\text{A}}=\tan5\text{A}$ Hence proved.

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