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How can we prove:sin5A-sin7A+sin8A-sin4A / cos4A+cos7A-cos5A-cos8A = cot6A?????

please answer!!!!!!!!

$To prove: \frac{\sin 5 A - \sin 7 A + \sin 8 A - \sin 4 A}{\cos 4 A+ \cos 7 A - \cos 5 A - \cos 8 A} = \cot 6 A Proof: L.H.S= \frac{(\sin 5 A - \sin 7 A) + (\sin 8 A - \sin 4 A)}{(\cos 4 A - \cos 8 A) + (\cos 7 A - \cos 5 A)} = \frac{2 \sin (\frac{- 2 A}{2}) \cos (\frac{12 A}{2}) + (2 \sin \frac{4 A}{2} \cos \frac{12 A}{2})}{- 2 \sin \frac{12 A}{2} \cdot \sin (\frac{- 4 A}{2}) + (- 2 \sin \frac{12 A}{2} \sin \frac{2 A}{2})} = \frac{2 \sin 2 A · \cos 6 A - 2 \sin A \cos 6 A}{2 \sin 6 A \cdot \sin 2 A - 2 \sin 6 A \cdot \sin A} = \frac{2 \cos 6 A [\sin 2 A - \sin A]}{2 \sin 6 A [\sin 2 A - \sin A]} = \frac{\cos 6 A}{\sin 6 A} = \cot 6 A = R.H.S [Hence proved]$

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