$\frac{cos 8 A cos 5 A - cos 12 A cos 9 A}{sin 8 A cos 5 A + cos 12 A sin 9 A} = tan 4 A$

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Solution

$\frac{cos 8 A cos 5 A - cos 12 A cos 9 A}{sin 8 A cos 5 A + cos 12 A sin 9 A} = \frac{\frac{1}{2} [(cos 13 A + cos 3 A) - (cos 21 A cos 3 A)]}{\frac{1}{2} [(sin 13 A + sin 3 A) + (sin 21 A sin 3 A)]} = \frac{cos 13 A - cos 21 A}{sin 13 A + sin 21 A} = \frac{2 sin (\frac{13 + 21}{2}) A sin (\frac{21 - 13}{2}) A}{2 sin (\frac{13 + 21}{2}) A cos (\frac{21 - 13}{2}) A} = \frac{sin 4 A}{cos 4 A} = tan 4 A$

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