## Verifying Identites

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

### Verifying Identites

Hey guys so I got a couple problems that I am stuck on and hoping to get some help

1.)( cot^2(x)-1)/(1+cot^x(x))=1-2sin^2(x) SOLVED

2.) sin^4(x)-cos^4(x)=2sin^2(x)-1 SOLVED

3.)tan^2(x)sin^2(x)=tan^2(x)+cos^2(x)-1 WORKING ON SEE BOTTOM POST 1

4.) ((sec(x)-tan(x))^2)/(sec(x)csc(x)-tan(x)csc(x))=2tan(x) WORKING ON SEE POST 2

5.) (csc(x)+cot(x))/(tan(x)+sin(x))=cot(x)csc(x) WORKING ON SEE POST 3

6.) sec^4(x)-sec^2(x)=tan^4(x)+tan^2(x) SOLVED

7.) (1+sin(x)+cos(x))^2=2(1+sin(x))(1+cos(x)) SOLVED

For some reason these were the only ones that stumped me however I will keep trying to chug em away while I wait for some help! Thanks guys
Last edited by Chyabrah on Sat Feb 18, 2012 8:59 pm, edited 2 times in total.
Chyabrah

Posts: 3
Joined: Sat Feb 18, 2012 3:57 am

Chyabrah wrote:Hey guys so I got a couple problems that I am stuck on

For instance, for the first equation, the left-hand side is clearly the more complicated, so you started working on that side. A standard technique is to convert things to sines and cosines, so you did that, and then converted to common denominators and simplified the complex fraction. What did you do next?

stapel_eliz

Posts: 1803
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Verifying Identites

Hey so I was able to solve the first 2 now working on number 3. I will do these one by one to clear confusion.

tan^2(x)sin^2(x)=tan^2(x)+cos^2(x)-1
3.) I started to work with the right hand side since it had an addition sign and looked ugly.

First, I converted (cos^2(x)-1) to -sin^2(x) so i had now
tan(x)^2-sin^2(x)

Second, I know tan=sin/cos so I looked at it as
sin^2(x)/cos^2(x)-sin^2(x)

Third, They dont have a like denominator so I multiplied sin -sin^2(x) by (cos^2(x)/cos^2(x))
(sin^2(x)-sin^2(x)cos^2(x))/cos^2(X)

Question: I just started thinking, can I factor out the sin^2(x) at the top in order to get sin^2(x)(1-cos^2(x))?

If so then I recognize that 1-cos^2(x) is also sin^2(x)..

However, all this typing up might have just made me solve it since I dont see why you couldn't factor out the sin^2(x)!
Chyabrah

Posts: 3
Joined: Sat Feb 18, 2012 3:57 am

### Re: Verifying Identites

((sec(x)-tan(x))^2+1)/(sec(x)csc(x)-tan(x)csc(x))=2tan(x)

This guy seems to be giving me a headache!

I started by simplifying the denominator.

(sec(x)csc(x)-tan(x)csc(x)) is equal to 1/(sin(x)cos(x))-sin(x)/(cos(x)/sin(x))

both have the same denominator so i simplified it to 1-sin(x)

---------------------------
Now on to the denominator

(sec(x)-tan(x))^2+1 is equal to (sec(x)-tan(x))(sec(x)-tan(x))+1

I then multiplied them to get
sec^2(x)+tan^2(x)-2tan(x)sec(x)+1

I now have (sec^2(x)+tan^2(x)-2tan(x)sec(x)+1)/(1-sin(x)) which is where I am stuck.
I then was able to simplify even more by making this sec^2(x)+sec^2(x)-2tan(x)sec(x)
Then factored out a sec^2(x) to get 2sec^2(x)((1-tan(x))

The next step I want to do is multiply by (1+sin(x)) so i can get (1-sin^2(x)) on the bottom which would make it cos^2(x)

Have I gone wrong somewhere or am I on the right track?
Chyabrah

Posts: 3
Joined: Sat Feb 18, 2012 3:57 am

### Re: Verifying Identites

You seem to be making good progress on identity (4). Once one arrives at this point with the numerator:

$2\sec^2(x)\left(1\, -\, \tan(x)\right)$

one may then convert to sines and cosines:

$\frac{2\, -\, 2\sin(x)}{\cos^2(x)}$

Then one may multiply this against the denominator as:

$\left(\frac{2\left(1\, -\, \sin(x)\right)}{\cos^2(x)}\right)\left(\frac{\sin(x)\cos(x)}{1\, -\, \sin(x)}\right)$

Cancel common factors, and simplify what remains.
nona.m.nona

Posts: 255
Joined: Sun Dec 14, 2008 11:07 pm