I have no idea how to solve this!
It's about Verifying trigonometric identities.
Here's the problem:

1 divided by 1+tanx = cotx divided by 1+cotx

Wouldn't it be easier to write:

1/(1+tan(x))=cot(x)/(1+cot(x))

Or, in LaTeX:

\frac{1}{1+tan(x)}=\frac{cot(x)}{1+cot(x)}

To see the code I used to make it display that way, click on 'quote' at the lower right corner of this post.

Just cross multiply:

1+cot(x)=cot(x)(1+tan(x))

Now, can you see it? What does

cot(x)\cdot tan(x) equal?

THank you !

So for another one, I have
4tanxcos²x-2tanx/1-tan²x=2tanx/1-tan²x.
So far, I have simplified it to 2tanxcos²x/1-tan²x.
Could I set up a proportion setting my identities equal to each other?

Are you allowed to cross multiply in proving identities :confused:

My method would be to divide both sides by tan, or cot, depending on where you are starting from.

\frac{1}{1 + \tan(x)} = \frac{1/\tan(x)}{(1 + \tan(x))/\tan(x)} = \frac{\cot(x)}{\cot(x) + 1} = \frac{\cot(x)}{1+ \cot(x)}


\frac{1}{1 + \cot(x)} = \frac{1/cot(x)}{(1 + \cot(x))/\cot(x)} = \frac{\tan(x)}{\tan(x) + 1} = \frac{\tan(x)}{1+ \tan(x)}

This is using the fact that \tan(x) = \frac{1}{\cot(x)}

~~~~~~~~~~~~~~~~
4\tan x\cos^2x-\frac{2\tan x}{1-\tan^2x}=\frac{2\tan x}{1-\tan^2x}

I don't think this is a true identity, because:

Multiply both sides by 1 - tan^2x gives a new identity:

4\tan x\cos^2x - 4 \tan^3x\cos^2x - 2\tan x= 2\tan x

Adding 2 tan x to give this new identity:

4\tan x\cos^2x - 4 \tan^3x\cos^2x = 4\tan x

Factor 4tan x on the left:

4\tan x(\cos^2x - \tan^2x\cos^2x) = 4\tan x

Simplifies to

4\tan x(\cos^2x - \sin^2x) = 4\tan x

And \cos^2x - \sin^2x = \cos(2x) \neq 1 as required for the identity to be true.

Either that, or you did a typing mistake, the identity should be instead:

4\tan x\cos^2x+\frac{2\tan x}{1-\tan^2x}=\frac{2\tan x}{1-\tan^2x}