L.H.S.

$$= \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}$$

Step 1: Group cos 4x and cos 2x in the numerator.

Using the identity $\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$,

$$\cos 4x + \cos 2x = 2 \cos\left(\frac{4x+2x}{2}\right) \cos\left(\frac{4x-2x}{2}\right) = 2 \cos 3x \cos x$$

Step 2: Therefore, numerator becomes:

$$\cos 4x + \cos 3x + \cos 2x = 2 \cos 3x \cos x + \cos 3x = \cos 3x\,(2 \cos x + 1)$$

Step 3: Similarly, group sin 4x and sin 2x in the denominator.

Using the identity $\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$,

$$\sin 4x + \sin 2x = 2 \sin\left(\frac{4x+2x}{2}\right) \cos\left(\frac{4x-2x}{2}\right) = 2 \sin 3x \cos x$$

Step 4: Therefore, denominator becomes:

$$\sin 4x + \sin 3x + \sin 2x = 2 \sin 3x \cos x + \sin 3x = \sin 3x\,(2 \cos x + 1)$$

Step 5: Now divide numerator by denominator:

$$\text{L.H.S.} = \frac{\cos 3x\,(2 \cos x + 1)}{\sin 3x\,(2 \cos x + 1)} = \frac{\cos 3x}{\sin 3x} = \textbf{cot 3x} = \text{R.H.S.}$$

∴ Hence Proved.