## Christians4 2 years ago Prove:

1. Christians4

$Tan^2x - \cos^2x = 1/\cos^2 - 1 - \cos^2x$

2. lgbasallote

can you rewrite that right side? is that $\frac 1{cos^2 x - 1} - \cos ^2 x$

3. Christians4

|dw:1352207602501:dw|

4. lgbasallote

you know this identity right? $\sin^2 x = 1 - \cos^2 x$

5. Christians4

yes

6. lgbasallote

wait..nevermind what i said....

7. Christians4

alright

8. lgbasallote

$\frac 1{\cos^2x} - 1 - \cos^2 x$ convert to common denominators first $\implies \frac 1{\cos^2 x} - \frac{\cos^2 x}{\cos^2 x} - \frac{\cos^4 x}{\cos^2 x}$ agree?

9. Christians4

yes :) I was on the right track whew :P

10. lgbasallote

now combine the numerators $\frac{1 - \cos^2 x - \cos ^ 4 x}{\cos^2 x}$ is this what you did?

11. Christians4

yes but Instead if making it cos^4x I cancelled out the top bracket Cos^2x with the bottom one...which is probably where I went wrong?

12. lgbasallote

no. that works too $\frac{1 - \cos^2 x}{\cos^2 x} - \cos^2 x$ yes?

13. Christians4

hmm, why didnt you just totally cancel out the denominator?

14. Christians4

I was left with 1-cos^2x-cos^2x :S

15. lgbasallote

ah. that's different

16. lgbasallote

you can't do that

17. HorribleAtMath

|dw:1352208030577:dw|

18. Christians4

hmm whats the reason I couldnt?

19. lgbasallote

observe what i did earlier $\frac 1{\cos^2 x} - 1 - \cos^2 x \implies \frac 1{\cos^2 x} - \frac{\cos^2 x}{\cos^2 x} - \frac{\cos^4 x}{\cos^2 x}$ each term has its own denominator

20. HorribleAtMath

|dw:1352208100756:dw|

21. HorribleAtMath

ap calc etcher tell me that

22. Christians4

OH I understand

23. lgbasallote

if you cancel out the cos^2 x below cos^4 x, you can only do it there, and not in the other terms. make sense?

24. Christians4

yes! I love you :P

25. lgbasallote

so anyway, the RHS simplifies to $\frac{1 - \cos^2 x}{\cos^2 x} - \cos^2 x$ i assume you know how to proceed form tehre?

26. Christians4

should i move to the left side?

27. Christians4

wo think i got it? L.S) will equel 1-cos^2/cos^2x - Cos^2x

28. Christians4

Ls = Rs Sweet! Thanks for your help! I also found a few more I am having trouble with...this next question has a tan^3x which I haven't used in a while i forget what i can do with it

29. lgbasallote

tan^3 x = tan^2 x * tan x

30. lgbasallote

that can work

31. lgbasallote

or maybe tan^3 x = sin^3 x / cos^3 x

32. Rosh007

$\tan ^{2}x - \cos^{2}x =\frac{1}{ \cos^2(x)} -1- \cos^{2}x$ $\tan ^{2}x - \cos^{2}x =\frac{(1-\cos^{2}x) }{ \cos^2(x)} - \cos^{2}x$ which is \$\tan ^{2}x =\frac{(1-\cos^{2}x) }{ \cos^2(x)}$ which is $\tan ^{2}x =\frac{(\sin^{2}x) }{ \cos^2(x)}$ which satisfis the relation

33. Christians4

Thanks guys, lemme try this other one and i cant get it ill post a new question

34. Rosh007

all the best :)