 Ask Remember Me? musicrulez Posts: 3, Reputation: 1 New Member #1 Jan 5, 2011, 11:43 PM
Precalculus
I have no idea how to solve this!
Here's the problem:

1 divided by 1+tanx = cotx divided by 1+cotx galactus Posts: 2,271, Reputation: 282 Ultra Member #2 Jan 6, 2011, 05:47 AM

Wouldn't it be easier to write:

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

Or, in LaTeX:

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:

Now, can you see it? What does

equal? musicrulez Posts: 3, Reputation: 1 New Member #3 Jan 11, 2011, 05:31 PM
Comment on galactus's post
THank you ! musicrulez Posts: 3, Reputation: 1 New Member #4 Jan 11, 2011, 05:36 PM
Comment on galactus's post
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? Unknown008 Posts: 8,076, Reputation: 723 Uber Member #5 Jan 11, 2011, 10:24 PM

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.

This is using the fact that

~~~~~~~~~~~~~~~~

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

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

Adding 2 tan x to give this new identity:

Factor 4tan x on the left:

Simplifies to

And as required for the identity to be true.

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

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