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Prove that: $\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$
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L.H.S. $\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x$
We have $3 x=x+2 x$
$\cot 3 x=\cot (x+2 x)$
$=\frac{\cot x \cot 2 x-1}{\cot x+\cot 2 x}$
By cross multiplication
$\cot 3 x(\cot x+\cot 2 x)=\cot x \cot 2 x-1$
$\cot x \cot 3 x+\cot 2 x \cot 3 x=\cos x \cot 2 x-1$
$\therefore \cos x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$
We have $3 x=x+2 x$
$\cot 3 x=\cot (x+2 x)$
$=\frac{\cot x \cot 2 x-1}{\cot x+\cot 2 x}$
By cross multiplication
$\cot 3 x(\cot x+\cot 2 x)=\cot x \cot 2 x-1$
$\cot x \cot 3 x+\cot 2 x \cot 3 x=\cos x \cot 2 x-1$
$\therefore \cos x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$
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