Question

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

Hint: Approach the solution by using $\cot (A + B) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}$ and prove L.H.S=R.H.S.

Given L.H.S part as
$\Rightarrow \cot x\cot 2x - \cot 2x\cot 3x - \cot 3x\cot x$
Let us take $\cot 3x$ term as common from last two terms
$\Rightarrow \cot x\cot 2x - \cot 3x(\cot 2x + \cot x)$
Here $\cot 3x$ can be written as $\cot 2x + \cot x$ where we can rewrite the above expression as
$\Rightarrow \cot x\cot 2x - (\cot 2x + \cot x)(\cot 2x + \cot x)$
Now let us apply $\cot (A + B) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}$ formula to one of the $\cot 2x + \cot x$ term.
On applying the formula and further simplification we get
$\Rightarrow \cot x\cot 2x - \left( {\dfrac{{\cot 2x\cot x - 1}}{{\cot x + \cot 2x}}} \right)(\cot 2x + \cot x)$
$\Rightarrow \cot x\cot 2x - (\cot 2x\cot x - 1) \\ \Rightarrow \cot x\cot 2x - \cot 2x\cot x + 1 \\ \Rightarrow 1 \\$
Hence we proved that L.H.S=R.H.S
Note: Focus on simplification after applying the formula. Try to take common terms out rearranged in a format so that we can apply formula easily.