1. ## False identity?

Hello, all!

I need to prove the following identity:

$\displaystyle \frac{cos(x-y)}{sin(x-y)} = \frac{1 + tan\hspace{5 pt}x\hspace{5 pt}tan\hspace{5 pt}y}{tan\hspace{5 pt}x + tan\hspace{5 pt}y}$

I am able to prove that the left side of the equation is equal to:

$\displaystyle \frac{1 + tan\hspace{5 pt}x\hspace{5 pt}tan\hspace{5 pt}y}{tan\hspace{5 pt} x - tan\hspace{5 pt}y}$

The obvious difference between what I am able to prove and what the identity actually says is the difference in sign between tan x and tan y in the denominator.

I'll be happy to explain how I arrived at my proof, but I'm hoping it should be pretty straight forward to understand.

So my question is this:

Is there some way to prove this identity, or is this a false identity (perhaps a misprint in the text I'm reading)?

Thank you!

2. How did you do it? I would expand the LHS.

3. Yes, I expanded the LHS using addition laws. I then multiplied both the numerator and denominator by:

$\displaystyle \frac{1}{cos\hspace{5pt}x\hspace{5pt}cos\hspace{5p t}y}$

That yields the answer I provided, with tan x - tan y in the denominator.

4. Hello, JennyFlowers!

I need to prove the following identity: .$\displaystyle \frac{\cos(x-y)}{\sin(x-y)} \;=\; \frac{1 + \tan x \tan y}{\tan x + \tan y}$ . Not true!

I am able to prove that the left side of the equation is equal to: .$\displaystyle \frac{1 + \tan x \tan y}{\tan x -\tan y}$ . You are right!

Is there some way to prove this identity, or perhaps a misprint in the text?
. . It's a misprint.

$\displaystyle \frac{\cos(x-y)}{\sin(x-y)} \;=\;\cot(x-y) \;=\;\frac{1}{\tan(x-y)} \;=\;\frac{1}{\dfrac{\tan x - \tan y}{1 + \tan x\tan y}} \;=\;\frac{1 + \tan x\tan y}{\tan x - \tan y}$

5. Thank you!