Question

Prove the following:
$\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x=1$

Answer 1

Hint:In the L.H.S of the expression, take $\cot 3x$ as common from $\cot 2x\cot 3x-\cot 3x\cot x$. After taking $\cot 3x$ as common, you will leave with $\cot 2x+\cot x$. Now, write $\cot 3x$ as $\cot \left( 2x+x \right)$ then use the identity of $\cot \left( A+B \right)$ on $\cot \left( 2x+x \right)$ then simplify.

Complete step-by-step answer:
We have to prove the following expression,
$\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x=1$
We are going to simplify the L.H.S of the above equation.
$\begin{aligned} & \cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x \\\ & =\cot x\cot 2x-\cot 3x(\cot 2x+\cot x) \\\ \end{aligned}$
Now, writing $\cot 3x$ as $\cot \left( 2x+x \right)$ in the above expression we get,
$\cot x\cot 2x-\cot (2x+x)(\cot 2x+\cot x)$
Now, applying $\cot \left( A+B \right)$ identity in $\cot \left( 2x+x \right)$ we get,
$\begin{aligned} & \cot (A+B)=\dfrac{\cot A\cot B-1}{\cot A+\cot B} \\\ & \cot (2x+x)=\dfrac{\cot 2x\cot x-1}{\cot 2x+\cot x} \\\ \end{aligned}$
$\begin{aligned} & \cot x\cot 2x-\cot (2x+x)(\cot 2x+\cot x) \\\ & =\cot x\cot 2x-\left( \dfrac{\cot 2x\cot x-1}{\cot 2x+\cot x} \right)(\cot 2x+\cot x) \\\ \end{aligned}$
From the above expression, you can see that $\cot 2x+\cot x$ will be cancelled out from the numerator and the denominator. So, the remaining expression will look like:
$\begin{aligned} & \cot x\cot 2x-\left( \cot 2x\cot x-1 \right) \\\ & =\cot x\cot 2x-\cot 2x\cot x+1 \\\ \end{aligned}$
In the above expression $\cot x\cot 2x$ will be cancelled out and the answer we get is:
1
The simplification of the L.H.S of the given expression yields 1 which is equal to R.H.S.
Hence, we have proved L.H.S = R.H.S of the given expression.

Note: The equation given in the question is:
$\cot x\cot 2x-\cot 2x\cot 3x-\cot 3x\cot x=1$
From the L.H.S of the above equation, we can also take $\cot 2x$ as common from$\cot x\cot 2x-\cot 2x\cot 3x$ and then write $\cot 2x$ as $\cot \left( 3x-x \right)$ and apply the identity of $\cot \left( A-B \right)$on $\cot \left( 3x-x \right)$ then also the simplification of L.H.S will give you 1.

& \cot 2x\left( \cot x-\cot 3x \right)-\cot 3x\cot x \\\ & =\cot (3x-x)\left( \cot x-\cot 3x \right)-\cot 3x\cot x \\\ & =\dfrac{1+\cot 3x\cot x}{\cot x-\cot 3x}\left( \cot x-\cot 3x \right)-\cot 3x\cot x \\\ \end{aligned}$$ In the above expression $\cot x-\cot 3x$ will be cancelled out and the remaining expression will look like: $\begin{aligned} & 1+\cot 3x\cot x-\cot 3x\cot x \\\ & =1 \\\ \end{aligned}$ Hence, we have simplified the L.H.S of the given expression to 1.