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:

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

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?
 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.

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

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

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:

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

 Question Tools Search this Question Search this Question: Advanced Search

## Check out some similar questions!

Proofing Pre Calculus Identities [ 3 Answers ]

prove the following identities? a.) (1 - cosX)(cscX + cotX)=sinX

Precalculus Word [ 3 Answers ]

Here's the problem: The growth rate y (in pounds per month) of infants is related to their present weight x (in pounds) by the formula y=cx(21-x) for some constant c>0. At what weight is the rate of growth a maximum. It says I need to use the -b/2a formula, but I'm not sure how to use that...

Precalculus - Identities [ 1 Answers ]

I'm stuck on a problem, and I want to figure it out so that I will know it later if it might appear on a test or something. cot2A = 1/2(cotA-tanA) Thanks!