Prove the following: (\frac{cos4x+cos3x+cos2x}{sin4x+sin3x+sin2x}=cot3x)

LHS=(\frac{cos4x+cos3x+cos2x}{sin4x+sin3x+sin2x})=(\frac{cos4x+cos2x+cos3x}{sin4x+sin2x+sin3x})=(\frac{2cos(\frac{4x+2x}{2})cos(\frac{4x-2x}{2})+cos3x}{2sin(\frac{4x+2x}{2})cos(\frac{4x-2x}{2})+sin3x})=cot3x=RHS
Hence proved