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.

Given $$\displaystyle \frac{\cot x}{1-\tan x}+\frac{\tan x}{1-\cot x} = \frac{1}{\sin x}\cdot \left(\frac{\cos^2 x}{\cos x-\sin x}\right) - \frac{1}{\cos x}\cdot \left(\frac{\sin^2 x}{\cos x-\sin x}\right)$$
So $$\displaystyle = \frac{\cos^3 x-\sin^3 x}{\sin x\cdot \cos x\cdot (\cos x-\sin x)} = \frac{(\cos x-\sin x)\cdot (\cos^2 x+\sin^2 x+\sin x\cdot \cos x)}{\sin x\cdot \cos x \cdot (\cos x-\sin x)}$$
So $$\displaystyle  = \frac{1+\sin x\cdot \cos x}{\sin x\cdot \cos x} = \sec x\cdot \csc x+1$$
So $$\displaystyle \frac{\cot x}{1-\tan x}+\frac{\tan x}{1-\cot x} = 1+\sec x\cdot \csc x$$
