Question : Simplify.
$\frac{\sin8\theta \cos\theta - \sin6\theta \cos3\theta}{\cos2\theta \cos\theta - \sin3\theta \sin4\theta}$
Option 1: $\cot \theta$
Option 2: $\cot 2 \theta$
Option 3: $\tan \theta$
Option 4: $\tan 2 \theta$
Correct Answer: $\tan 2 \theta$
Solution :
Given:
$\frac{\sin8\theta \cos\theta - \sin6\theta \cos3\theta}{\cos2\theta \cos\theta - \sin3\theta \sin4\theta}$
= $\frac{2\sin8\theta \cos\theta - 2\sin6\theta \cos3\theta}{2\cos2\theta \cos\theta - 2\sin3\theta \sin4\theta}$
= $\frac{\sin(8\theta+\theta)+\sin(8\theta-\theta)-\sin(6\theta+3\theta)-\sin(6\theta-3\theta)}{\cos(2\theta+\theta)+\cos(2\theta-\theta)-\cos(3\theta-4\theta)+\cos(3\theta+4\theta)}$
= $\frac{\sin9\theta + \sin7\theta - \sin9\theta - \sin3\theta}{\cos3\theta + \cos\theta - \cos(-\theta)+cos7\theta}$
= $\frac{\sin7\theta-\sin3\theta}{\cos3\theta + \cos\theta - \cos\theta+cos7\theta}$
= $\frac{\sin7\theta-\sin3\theta}{\cos7\theta+cos3\theta}$
= $\frac{2\cos\frac{7\theta+3\theta}{2}\sin\frac{7\theta-3\theta}{2}}{2\cos\frac{7\theta+3\theta}{2}\cos\frac{7\theta-3\theta}{2}}$
= $\frac{\cos5\theta\sin2\theta}{\cos5\theta\cos2\theta}$
= $\frac{\sin2\theta}{\cos2\theta}$
= $\tan2\theta$
Hence, the correct answer is $\tan2\theta$.
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