# Thread: Help proving one side equals the other.

1. ## Help proving one side equals the other.

The original equation is: " 1/1-cosx - 1/ 1+secx = csc^2x + cot^2x "

These are the steps i've used so far:

= 1/1-cosx (1+cosx/1+cosx) - 1/1+secx (1-secx/1-secx) ---multiply by the conjugate

= 1+cosz/sin^2x + 1-secx/tan^2x

Thats as far as i've been able to get.

Any help would be greatly appreciated!

2. \begin{aligned}
\frac{1}{1-\cos x}-\frac{1}{1+\sec x}&=\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} \\
& =\frac{1+\cos x-\cos x+\cos ^{2}x}{1-\cos ^{2}x} \\
& =\frac{1}{\sin ^{2}x}+\frac{\cos ^{2}x}{\sin ^{2}x}.
\end{aligned}

You may conclude now.

3. Thanks for the quick response!!

Is there any way that you could expand your work a bit, im having trouble following it. I know that secx = 1/cosx, but how do you get that to where it is cosx/1+cosx, and after that im confused a bit as well.

Thanks!

4. No problem.

We have $\frac{1}{1+\sec x}.$ But you know that $\cos x\sec x=1,$ then you just multiply top & bottom by $\cos x,$ and the result follows.

5. ok, when you multiply the left side by cosx/cosx, i see how that would give you cosx as the numerator, but wouldnt that leave you with 1+secxcosx as the denominator, how does that eqaul 1+cosx? And after that step to get 1/1-cosx - cosx/1+cosx are you multiplying both sides by their conjugate to get a common denominator so that you can combine them to 1-cos^2x?
Thanks for all the help, i just learned this a few days ago and am still trying to understand all the transformations and steps involved.

6. Originally Posted by masteroc

ok, when you multiply the left side by cosx/cosx, i see how that would give you cosx as the numerator, but wouldnt that leave you with 1+secxcosx as the denominator, how does that eqaul 1+cosx?
Read post #4. (After multiplication yields $\cos x+\cos x\sec x.$)

Originally Posted by masteroc

And after that step to get 1/1-cosx - cosx/1+cosx are you multiplying both sides by their conjugate to get a common denominator so that you can combine them to 1-cos^2x?
Yes.

7. Ah, i got it now!

Thanks for bearing with me on that. As I said, I just learned this and im still trying to get used to all the transformations and steps!

Thanks again Krizalid!