QPaperGen

We use the identity:
\[
\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}
\]

Now, consider the left-hand side (LHS):
\begin{align*}
\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x
&= \cot x \cot 2x - \cot 3x (\cot 2x + \cot x) \\
&= \cot x \cot 2x - \cot(2x + x)(\cot 2x + \cot x) \\
&= \cot x \cot 2x - \left[ \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x} \right](\cot 2x + \cot x) \\
&= \cot x \cot 2x - (\cot 2x \cot x - 1) \\
&= 1 \qquad \textbf{(Hence, proved.)}
\end{align*}