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$\cot x.\cot 2x - \cot 2x.\cot 3x - \cot x.\cot 3x = 1$

Answer
Verified

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

Complete step-by-step answer:

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.

Complete step-by-step answer:

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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