How to proceed from $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$ To prove: $\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x) = 1$
My attempt at the solution:
\begin{gather}\frac{\cos(x)\cos(2x)}{\sin(x)\sin(2x)}-\frac{\cos(2x)\cos(3x)}{\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{\cos(x)\cos(2x)\sin(3x)-\cos(2x)\cos(3x)\sin(x)}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{ \cos(2x)[ \cos(x)\sin(3x)-\cos(3x)\sin(x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{\cos(2x)[\sin(4x)\sin(2x)-\cos(3x)\sin(x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{\cos(2x)[2\sin(4x)\sin(2x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{2\cos(2x)\sin(4x)}{\sin(x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}\\\\
\frac{2\cos(2x)\sin(4x)}{\sin(x)\sin(3x)}-\frac{\cos(4x)\cos(2x)}{2\sin(3x)\sin(x)}\end{gather}
The problem is, I don't know where to go from here (and due to so many calculations involved, I'm also not sure of the above result).
Also, if you see a more elegant way to solve this, please provide a hint (not the complete solution).
 A: Do it this way- Expand $\cot(3x-2x-x)$ in the 
$$ \cot(A+B+C)  = \dfrac{\cot(A)+\cot(B)+\cot(C)-3\cot(A)\cot(B)\cot(C)}{ 1-\cot(A)\cot(B)-\cot(B)\cot(C)-\cot(C)\cot(A)}$$
We know $\cot(0) = \infty$. The denominator is zero.So, ...you got it already.
A: Hints:
(1)  Factor out $$\frac{\cos(2x)}{\sin(x)\sin(3x)}$$ from your last expression.
(2)  Simplify $$2\sin(4x) - \frac{1}{2}\cos(4x).$$
Can you take it from here?
A: $$\cot(A+B)=\dfrac{1-\tan A\tan B}{\tan A+\tan B}=\dfrac{\cot A\cot B-1}{\cot B+\cot A}$$
$$\iff\cot A\cot B=1+\cot(A+B)[\cot B+\cot A]$$
Set $A=x, B=2x$
$$\cot(A-B)=\dfrac{1+\tan A\tan B}{\tan A-\tan B}=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$$
$$\iff\cot A\cot B=\cot(A-B)[\cot B-\cot A]-1$$
Set $A=3x,B=2x$
and  $A=3x,B=x$
A: Notice, here is right approach followed by OP.,  $$LHS=\cot(x)\cot(2x)-\cot(2x)\cot(3x)-\cot(3x)\cot(x)$$
$$=\frac{\cos(x)\cos(2x)}{\sin(x)\sin(2x)}-\frac{\cos(2x)\cos(3x)}{\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
$$=\frac{\cos(2x)[\sin(3x)\cos(x)-\sin(x)\cos(3x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
using $\color{blue}{\sin A\cos B-\sin B\cos A=\sin(A-B)}$, 
$$=\frac{\cos(2x)[\sin(3x-x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
$$=\frac{\cos(2x)}{\sin(x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
$$=\frac{\cos(2x)-\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
$$=\frac{\cos(3x-x)-\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
using $\color{blue}{\cos(A-B)=\cos A \cos B+\sin A\sin B}$, 
$$=\frac{\cos(3x)\cos(x)+\sin(3x)\sin(x)-\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
$$=\frac{\sin(3x)\sin(x)}{\sin(3x)\sin(x)}$$
$$=1=RHS$$
A: Its surprising that everyone else missed this out, but there's actually a very simple and elegant solution to this proof.
So although this question is more than a year old, I still decided to post my proof.

I begin with a simple equation:
$$\cot(2x+x)=\cot{3x}$$

Now applying the formula: $\cot(A+B)=\frac{\cot{A}\cot{B}-1}{\cot{A}+\cot{B}}$, we get:

$$\frac{\cot{2x}\cot{x}-1}{\cot{2x}+\cot{x}}=\cot{3x}$$
$$\cot{2x}\cot{x}-1=\cot{2x}\cot{3x}+\cot{3x}\cot{x}$$
$$\cot{x}\cot{2x}-\cot{2x}\cot{3x}-\cot{3x}\cot{x}=1$$

and you're done.
A: Go slowly:
\begin{align}
\cot x\cot2x-\cot2x\cot3x
&=
\cot2x\left(\frac{\cos x}{\sin x}-\frac{\cos3x}{\sin3x}\right)\\[6px]
&=\frac{\cos2x}{\sin2x}\frac{\sin3x\cos x-\cos3x\sin x}{\sin x\sin 3x}
\\[6px]
&=\frac{\cos2x}{\sin2x}\frac{\sin2x}{\sin x\sin 3x}\\[6px]
&=\frac{\cos2x}{\sin x\sin 3x}
\end{align}
So you want to compute
$$
\frac{\cos2x}{\sin x\sin 3x}-\frac{\cos3x}{\sin3x}\frac{\cos x}{\sin x}
=\frac{\cos2x-\cos3x\cos x}{\sin x\sin3x}
$$
Now
$$
\cos2x=\cos(3x-x)=\cos3x\cos x+\sin3x\sin x
$$
and you are done.
A: Here are your work upto the line I considered a problem occured:
$$\frac{\cos(x)\cos(2x)}{\sin(x)\sin(2x)}-\frac{\cos(2x)\cos(3x)}{\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
$$\frac{\cos(x)\cos(2x)\sin(3x)-\cos(2x)\cos(3x)\sin(x)}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
$$\frac{ \cos(2x)[ \cos(x)\sin(3x)-\cos(3x)\sin(x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
This line is incorrect to me.
$$\frac{\cos(2x)[\sin(4x)\sin(2x)-\cos(3x)\sin(x)]}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
Using $\sin(2x) = \sin(3x -x) = \sin(3x)\cos(x) - \sin(x)\cos(3x)$, It should be 
$$\frac{\cos(2x)\sin(2x)}{\sin(x)\sin(2x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
Then cancle $sin2x$
$$\frac{\cos(2x)}{\sin(x)\sin(3x)}-\frac{\cos(3x)\cos(x)}{\sin(3x)\sin(x)}$$
One more merge:
$$\frac{\cos(2x) - \cos(3x)\cos(x)}{\sin(x)\sin(3x)}$$
Using $\cos(2x) = \cos(3x -x) = \cos(3x)\cos(x) + \sin(3x)\sin(x)$, the last one become
$$\frac{\sin(3x)\sin(x)}{\sin(3x)\sin(x)} = 1$$
