# Help needed with Verifying Trigonometric Functions

1. Mar 5, 2009

### amd123

1. The problem statement, all variables and given/known data

$$\frac{secx-cscx}{secx+cscx}=\frac{tanx-1}{tanx+1}$$

$$\frac{(tan^{2}x - cot^{2}x)}{(tanx + cotx)}= (tanx - cotx)$$

$$tan^{2}2x+sin^{2}2x+cos^{2}2x=sec^{2}2x$$

$$cot^{2}2x+cos^{2}2x+sin^{2}2x=csc^{2}2x$$

3. The attempt at a solution
I've tried many times in my notebook and i'm posting these on hear as a last resort. Could someone please explain how these are done as my teacher LACKS the ability to teach.

2. Mar 5, 2009

### sutupidmath

i am assuming you are trying to prove these identities, right?

Well, show us what you have done so far. For the first one LHS, try to convert all of them into sin and cos, and after that see if you will get any tan's in there.

another helpful formula for the second one is

$$a^2-b^2=(a-b)(a+b)$$

Another helpful identity, which you shoul know, is:

$$sin^2(H)+cos^2(H)=1$$

Combine all these, and you will do fine!

3. Mar 5, 2009

### amd123

http://img131.imageshack.us/img131/8479/scan0004d.jpg [Broken]
http://img12.imageshack.us/img12/9265/scan0005o.jpg [Broken]
sorry that took so long my scanner sometimes doesn't want to scan stuff thats not in dark ink

Last edited by a moderator: May 4, 2017
4. Mar 5, 2009

### sutupidmath

$$\frac{(tan^{2}x - cot^{2}x)}{(tanx + cotx)}= (tanx - cotx)$$

now what u need to do is

tan^2x-cot^2x=(tanx-cotx)(tanx+cotx), i already provided u with this formula.
Now only simplify out.

On the 3rd and 4th use the fact that

$$sin^2(2x)+cos^2(2x)=1$$ and also the def. of tan. in terms of cos and sine.

5. Mar 5, 2009

### sutupidmath

$$tan^22x+1=\frac{sin^22x}{cos^22x}+1=..$$ find the common denominator and then u are done

Do the same thing for cot.

6. Mar 5, 2009

### amd123

thanks for the help :D