Question

 Prove the following expression, cot 2xcot x-cot 3xcot 2x-cot 3xcot x=1.

Answer 1

Hint: To prove the desired relation, we should know some trigonometric identities like, $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\left( \tan a \right)\left( \tan b \right)}$. We should also know that $\tan \theta =\dfrac{1}{\cot \theta }$. By using these identities, we can prove the desired relation given in the question.

Complete step-by-step answer:
In this question, we have been asked to prove that, $\Rightarrow\cot 2x\cot x-\cot 3x\cot 2x-\cot 3x\cot x=1$. To prove this, we will first start with the very basic algebraic addition property, which is, $x+2x=3x$. We know that if the angles are equal, then their tangent ratios are also equal, so we get, $\Rightarrow\tan \left( x+2x \right)=\tan 3x$. We also know that $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\left( \tan a \right)\left( \tan b \right)}$. So, by substituting this in the expression, $\tan \left( x+2x \right)=\tan 3x$, we can write this expression as,
$\dfrac{\tan x+\tan 2x}{1-\tan x\tan 2x}=\tan 3x$
Now, we will cross multiply the above equation. By doing so, we get, $\tan x+\tan 2x=\tan 3x\left( 1-\tan x\tan 2x \right)$
We can further simplify and write it as,
$\tan x+\tan 2x=\tan 3x-\tan x\tan 2x\tan 3x$
We will take $\tan 3x$ to the left hand side. By doing so, we can write the above equation as,
$\Rightarrow\tan x+\tan 2x-\tan 3x=-\tan x\tan 2x\tan 3x$
As we know that $\tan \theta =\dfrac{1}{\cot \theta }$, we will apply that in the above expression and write it as,
$\Rightarrow\dfrac{1}{\cot x}+\dfrac{1}{\cot 2x}-\dfrac{1}{\cot 3x}=-\dfrac{1}{\cot x\cot 2x\cot 3x}$
Now, we will take the LCM on the left hand side. By doing so, we will get the above equation as,
$\Rightarrow\dfrac{\left( \cot 2x \right)\left( \cot 3x \right)+\left( \cot x \right)\left( \cot 3x \right)-\left( \cot x \right)\left( \cot 2x \right)}{\left( \cot x \right)\left( \cot 2x \right)\left( \cot 3x \right)}=\dfrac{-1}{\left( \cot x \right)\left( \cot 2x \right)\left( \cot 3x \right)}$
Now, we can see that $\left( \cot x \right)\left( \cot 2x \right)\left( \cot 3x \right)$ is common in the denominator of both the sides, so we can cancel them. By cancelling them we get the equation as,
$\Rightarrow\cot 2x\cot 3x+\cot x\cot 3x-\cot x\cot 2x=-1$
Now, we will multiply both the sides by $\left( -1 \right)$. By doing so, we get,
$-\cot 2x\cot 3x-\cot x\cot 3x+\cot x\cot 2x=1$
Or we can write the above equation as,
$\cot 2x\cot x-\cot 3x\cot 2x-\cot 3x\cot x=1$
Which is the expression that we have been asked to prove in the question.
Hence, we have proved that, $\cot 2x\cot x-\cot 3x\cot 2x-\cot 3x\cot x=1$.

Note: The possible mistakes that students make while solving this question is by simplifying the equation from either the right hand side or the left hand side. This is not wrong, but they might get stuck after a certain point and will be wasting the time in proving the desired result.