## anonymous one year ago Trig / Pre Cal/ identities Guide me please. $$\frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x} + 1$$

1. anonymous

It is suppose to be. $$\frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x}$$

2. anonymous

$$\large \rm \frac{\frac{\sin x}{\cos x}}{\frac{1}{cos x}+1 }= \frac{\sec x+1}{\tan x}$$

3. anonymous

These things are killing me lol

4. anonymous

not an identity

5. xapproachesinfinity

okay so we have $\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1}=\frac{\sec x+1}{\tan x }$

6. anonymous

I got that they're reciprocals

7. xapproachesinfinity

if so we can do this $\Huge \frac{\tan x}{\sec x+1}=\frac{\sec x+1}{\tan x}$ clearly this is not truee

8. xapproachesinfinity

try some vlues if you have doubt

9. anonymous

Yes but I am trying to get from the left side to the right. $$\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1}=\frac{\sec x+1}{\tan x }$$ Do I times by cos?? $$\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1} \times cos x$$

10. anonymous

The identity is false is what he's saying. You won't be able to get from the left to the right.

11. xapproachesinfinity

now it does not work! left does not equal right

12. xapproachesinfinity

plug in some values to be sure if you have doubt

13. xapproachesinfinity

no* for (now)

14. anonymous

The original problem is $$\Huge \frac{tan x}{\sec - 1 }=\frac{\sec x+1}{\tan x }$$

15. xapproachesinfinity

0 for example is not good left give you 0 right gives you undefined 2/0 if it is an identity it should not be that way

16. anonymous

I did the left side assuming that is what we have to do.

17. xapproachesinfinity

is the left 1/cosx -1 or 1/cosx +1 you just changed in your last reply?

18. xapproachesinfinity

in that case it is an identity

19. xapproachesinfinity

see that you give us the wrong thing from the start

20. xapproachesinfinity

if it is a minus on the bottom it is an identity since tan^2x=sec^2x-1

21. anonymous

This is the original problem. $$\Huge \frac{tan x}{\sec - 1 }=\frac{\sec x+1}{\tan x }$$ How do we know to work the numerator or the denominator or both?

22. anonymous

For instance, should I $$\Huge \frac{tan x}{\frac{1}{cos x}-1 }=\frac{\sec x+1}{\tan x }$$ OR $$\Huge \frac{\frac{sin}{cos}}{{\sec x}-1 }=\frac{\sec x+1}{\tan x }$$ OR $$\Huge \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1}$$

23. xapproachesinfinity

usually when you prove something you should get from left to right of vice versa so taking left $\frac{\tan x}{\sec x=1}=\frac{\tan x(\sec x+1)}{(\sec x-1)(\sec x+1)}$ $=\frac{\tan x(\sec x+1)}{\tan^2x }=\frac{\sec x+1}{\tan x} ~~~Q.E.D$

24. xapproachesinfinity

you didn't need to do all that just leave tan and sec that way

25. xapproachesinfinity

you are making it hard

26. anonymous

There are no rules though

27. xapproachesinfinity

what rules?

28. xapproachesinfinity

did you get what i did?

29. anonymous

No

30. xapproachesinfinity

if we do it your way we have $\frac{\sin x}{1-\cos x}=\frac{\sin x(1+\cos x)}{1-\cos^2x}=\frac{1+\cos x}{\sin x}$ multiplyinh by 1/cos top and bottom we get $=\frac{\sec x+1}{\tan x}$

31. xapproachesinfinity

perhaps i'm going fast?

32. xapproachesinfinity

i skip some stuff given that you know how to deal with algebra

33. anonymous

I don't get it. How do I know to work the numerator or the denominator or both. How do I know what method to use when there are no rules to guy you.

34. anonymous

For instance, how did you do it your way?

35. xapproachesinfinity

like $\huge \frac{\frac{a}{b}}{\frac{1}{b}-1}=\frac{a}{b(\frac{1}{b}-1)}=\frac{a}{1-b}$

36. xapproachesinfinity

this is what i did with your way changing tan to sin/cos and sec to 1/cos

37. anonymous

Yes that is my way.

38. anonymous

But there are other ways and some ways don't work

39. xapproachesinfinity

yes once you get sin x/1-cosx you multiply by 1+cosx top and bottom

40. anonymous

Correct

41. xapproachesinfinity

what other Ways do you mean if you are using identities they should work

42. xapproachesinfinity

my way is no different than yours i still had to multiply top and bottom be 1+sec x to get sec^2x-1 on the bottom which is the same as tan^2x

43. xapproachesinfinity

this is just a matter of knowing your identities and how to use them

44. anonymous

For instance $$\Huge \frac{tan x}{\frac{1}{cos x}-1 }=\frac{\sec x+1}{\tan x }$$ $$\Huge \frac{(cosx)(tan x)}{cos x(\frac{1}{cos x}-1) }=\frac{\sec x+1}{\tan x }$$

45. xapproachesinfinity

it is still working since top becomes sinx

46. xapproachesinfinity

gotta go! i think you got this now?

47. anonymous

Ok so we have $$\Huge \frac{(cosx)(tan x)}{cos x(\frac{1}{cos x}-1) }=\frac{\sec x+1}{\tan x }$$ $$\Huge \frac{sin x}{1-cos x}=\frac{\sec x+1}{\tan x }$$

48. xapproachesinfinity

yes!

49. xapproachesinfinity

i explained above how you obtain the right side