ii) sin 8A cos A -cos 3A sin 6A / cos 2A cos A - sin 3A sin 4A = tan 2A

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Solution

$\frac{\sin (8 A) \cos (A) - \cos (3 A) \sin (6 A)}{\cos (2 A) \cos (A) - \sin (3 A) \sin (4 A)} = \frac{2 \sin (8 A) \cos (A) - 2 \cos (3 A) \sin (6 A)}{2 \cos (2 A) \cos (A) - 2 \sin (3 A) \sin (4 A)} = \frac{[\sin (8 A + A) + \sin (8 A - A)] - [\sin (6 A + 3 A) + \sin (6 A - 3 A)]}{[\cos (2 A + A) + \cos (2 A - A)] - [\cos (4 A - 3 A) - \cos (4 A + 3 A)]} = \frac{\sin (9 A) + \sin (7 A) - \sin (9 A) - \sin (3 A)}{\cos (3 A) + \cos (A) - \cos (A) + \cos (7 A)} = \frac{\sin (7 A) - \sin (3 A)}{\cos (7 A) + \cos (3 A)} = \frac{2 \cos (\frac{7 A + 3 A}{2}) \sin (\frac{7 A - 3 A}{2})}{2 \cos (\frac{7 A + 3 A}{2}) \cos (\frac{7 A - 3 A}{2})} = \frac{\cos (5 A) \sin (2 A)}{\cos (5 A) \cos (2 A)} = \tan (2 A) = RHS$

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