Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$.

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(A) $L.H.S. = \cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x$
$= \cot x \cot 2x - \cot 3x(\cot 2x + \cot x)$
$= \cot x \cot 2x - \cot(2x + x)(\cot 2x + \cot x)$
$= \cot x \cot 2x - \left[ \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x} \right](\cot 2x + \cot x)$
$\left[ \because \cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \right]$
$= \cot x \cot 2x - (\cot 2x \cot x - 1) = 1 = R.H.S.$

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