Thread: 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!

You may conclude now.

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!

No problem.

We have But you know that then you just multiply top & bottom by and the result follows.

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.

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 )

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.

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!