QPaperGen

We use the standard identities:


\[
\cos A + \cos B = 2 \cos\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]




\[
\sin A + \sin B = 2 \sin\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]



Now, consider the left-hand side (LHS):


\[
\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}
= \frac{(\cos 4x + \cos 2x) + \cos 3x}{(\sin 4x + \sin 2x) + \sin 3x}
\]



Apply the sum formulas:


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





\[
= \frac{2 \cos(3x) \cos(x) + \cos(3x)}{2 \sin(3x) \cos(x) + \sin(3x)}
= \frac{\cos(3x)(2 \cos x + 1)}{\sin(3x)(2 \cos x + 1)}
\]





\[
= \frac{\cos(3x)}{\sin(3x)} = \cot(3x)
\]



Therefore,


\[
\text{LHS} = \text{RHS}
\quad \boxed{\text{Hence proved.}}
\]