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# Prove the equation given below$\cot x.\cot 2x - \cot 2x.\cot 3x - \cot x.\cot 3x = 1$  Hint- Start with taking out cot$3x$ common from the last two terms and then write $3x$ as $2x+x$ , now use the identity $\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}$ .

Given equation is $\cot x.\cot 2x - \cot 2x.\cot 3x - \cot x.\cot 3x = 1$
By taking L.H.S we have
$= \cot x.\cot 2x - \cot 2x.\cot 3x - \cot x.\cot 3x$
Take $\cot 3x$ common from above term.
$\cot x.\cot 2x - \cot 3x\left[ {\cot 2x + \cot x} \right]$
Here we can write $\cot 3x{\text{ as }}\cot \left( {x + 2x} \right)$
Therefore the term becomes
$= \cot x.\cot 2x - \cot \left( {x + 2x} \right)\left[ {\cot 2x + \cot x} \right]$
AS we know that $\left[ {\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}} \right]$
By applying this formula, we get
$\Rightarrow \cot x\cot 2x - \left( {\dfrac{{\cot 2x\cot x - 1}}{{\cot 2x + \cot x}}} \right)\left( {\cot 2x + \cot x} \right)$
Now by simplifying the above term, we get
$\Rightarrow \cot 2x\cot x - \left( {\cot 2x\cot x - 1} \right) \\ \Rightarrow \cot 2x\cot x - \cot 2x\cot x + 1 \\ \Rightarrow 1 \\$
Hence, the result is same as R.H.S

Note- In order to solve these types of problems, first of all remember all the trigonometric identities and formulas. In the above question we use a formula of cotangent to split its angles. Similarly remember formulas to change product into sum and sum into product.

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