Question

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

Answer 1

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