# 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.

• Please read this tutorial on how to format mathematics on this site. – N. F. Taussig Aug 13 '15 at 9:48

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}

$\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot(x)} - 1 = \sec x \csc x$
=$\frac1{\tan x(1-\tan x)}-\frac{\tan^2 x}{1-\tan x}-1$ Because $(\tan x=\frac{1}{\cot x})$;Change $+$ to $-$.
=$\frac1{1-\tan x}(\frac{1-\tan^3 x}{\tan x})-1$ Take $1-\tan x$ common.
=$\frac{\tan^2 x+\tan x+1-\tan x}{\tan x}$
=$\frac{\sec^2 x}{\tan x}$
=$\frac{\frac1{\cos x}}{\sin x}$
=$\sec x\csc x$
Q.E.D

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 $$\displaystyle \frac{\cot x}{1-\tan x}+\frac{\tan x}{1-\cot x} = 1+\sec x\cdot \csc x$$