Prove that $\frac{cos 2 A cos 3 A - cos 2 A cos 7 A + cos A cos 10 A}{sin 4 A sin 3 A - sin 2 A sin 5 A + sin 4 A sin 7 A} = cot 6 A cot 5 A$

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Solution

Multiply above and below by $2$ and apply
LHS $\frac{(cos 5 A + cos A) - (cos 9 A + cos 5 A) + (cos 11 A + cos 9 A)}{(cos A + cos 7 A) - (cos 3 A + cos 7 A) + (cos 3 A + cos 11 A)} = \frac{(cos A + cos 11 A)}{(cos A - cos 11 A)} = \frac{2 cos 6 A cos 5 A}{2 sin 6 A sin 5 A} = cot 6 A cot 5 A$

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