How to prove the trigonometric identity $\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot x} - 1 = \sec x \csc x$ I am doing some practice questions for a Math class and I was told that similar questions would be in the exam. So I need to learn this but I have no idea where to even start with this question:
$$\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot x} - 1 = \sec x \csc x$$
Hint: use standard factorization for the difference of 2 cubes, e.g. $$a^3-b^3 = (a-b)(a^2 + ab + b^2)$$
Help please I find that looking at the working when someone does these questions allows me to learn the method. Thanks in advance.

Start from the left-hand side:
\begin{align}
\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot x} - 1
&=\frac{\dfrac{\cos x}{\sin x}}{1-\dfrac{\sin x}{\cos x}}
 +\frac{\dfrac{\sin x}{\cos x}}{1-\dfrac{\cos x}{\sin x}}-1\\
&=\frac{\cos^2x}{\sin x(\cos x-\sin x)}
 -\frac{\sin^2x}{\cos x(\cos x-\sin x)}-1\\
&=\frac{\cos^3x-\sin^3x}{\sin x\cos x(\cos x-\sin x)}-1\\
&=\frac{\cos^2x+\sin x\cos x+\sin^2x}{\sin x\cos x}-1\\
&=\frac{1}{\sin x\cos x}
\end{align}
