Proof of the trigonometric identity $\frac{1 + \tan x}{1 + \cot x} = \frac{1 - \tan x}{\cot x - 1}$

Question

I'm stuck on the equation $$\frac{1 + \tan x}{1 + \cot x} = \frac{1 - \tan x}{\cot x - 1}$$ I've tried everything I could think of to solve it but nothing is working. Where do I start?

Answer 1

The identity is not true. For $x \in (0,\frac {\pi} 4)$ LHS is positive and RHS is negative .

Answer for the revised version: cross multiply; it is fairly easy to verify that $(1+\tan x) (\cot x -1)=(1+\cot x) (1-\tan x)$. The only thing you need is the fact that $\tan x \cot x=1$.

Answer 2

For non-zero finite $\tan x,$

$$\dfrac{1+\tan x}{1-\tan x}=\dfrac{1+\dfrac1{\cot x}}{1-\dfrac1{\cot x}}=?$$

Now rearrange

Answer 3

Setting $\boxed{t = \tan x}$ and noting that $\boxed{\cot x = \frac{1}{t}}$ you have

$$t = t\frac{1-t}{1-t} = \frac{1-t}{\frac{1}{t}-1} = \frac{1-\tan x}{\cot x - 1}$$

Similarly, $$t = t\frac{1+t}{1+t} = \frac{1+t}{\frac{1}{t}+1} = \frac{1+\tan x}{1+\cot x}$$