Question:
Prove that $\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$
Solution:
L.H.S. $=\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x$
$=\cot x \cot 2 x-\cot 3 x(\cot 2 x+\cot x)$
$=\cot x \cot 2 x-\cot (2 x+x)(\cot 2 x+\cot x)$
$=\cot x \cot 2 x-\left[\frac{\cot 2 x \cot x-1}{\cot x+\cot 2 x}\right](\cot 2 x+\cot x)$
$\left[\because \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}\right]$
$=\cot x \cot 2 x-(\cot 2 x \cot x-1)$
$=1=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S}$