Question

Prove the equation given below
$\cot x.\cot 2x - \cot 2x.\cot 3x - \cot x.\cot 3x = 1$

Answer 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 \[\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}\] .

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.