## pottersheep Group Title Please help me prove this trig identity.... (secx - cosecx)/(tanx - cotx) = (tanx - cotx) /(secx - cosecx) 2 years ago 2 years ago

1. myininaya Group Title

i would write both sides in terms of sin(x) and cos(x) and clear the compound fractions

2. myininaya Group Title

and then you might have to use some other identities

3. moses_ Group Title

break up the LHS into $\frac{sec x}{tan x - cot x} + \frac{-csc x}{tan x - cot x}$

4. pottersheep Group Title

Oh....I'll try it that way then but Im not sure if I can reach the answer still.... And how can I change them to cos and sin? they are not squared :(

5. moses_ Group Title

my advice would be to multiply by one using $\frac{sinx}{sinx}$

6. moses_ Group Title

he means turn sec x into $\frac{1}{cos x}$ etc...

7. moses_ Group Title

for multiplying by one like i said above, dont use sin...use something that will reduce everything...

8. pottersheep Group Title

hmmm

9. pottersheep Group Title

i tried turning tan into sin/cos and cot into cis/sin and then I got all tans and cos's too

10. bluebrandon Group Title

try taking everything in terms of cos and sin and then simplifying as much as you can. Don't leave any sec, csc, tan or cot

11. Mertsj Group Title

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12. myininaya Group Title

i'm a her $\frac{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}}=\frac{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}}$ now for both sides we want to clear the compound fraction action going on $\frac{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}} \cdot \frac{\cos(x)\sin(x)}{\cos(x)\sin(x)}=\frac{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}} \cdot \frac{\cos(x)\sin(x)}{\cos(x)\sin(x)}$ $\frac{\sin(x)-\cos(x)}{\sin^2(x)-\cos^2(x)}=\frac{\sin^2(x)-\cos^2(x)}{\sin(x)-\cos(x)}$ now we can some factoring on both sides $\frac{\sin(x)-\cos(x)}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))}=\frac{(\sin(x)+\cos(x))(\sin(x)-\cos(x))}{\sin(x)-\cos(x)}$ so we have $\frac{1}{\sin(x)+\cos(x)}=\sin(x)+\cos(x)$ ...

13. Mertsj Group Title

Now. Use the same technique on the right side.

14. myininaya Group Title

this doesn't look like an identity

15. moses_ Group Title

oh my apologies

16. pottersheep Group Title

Wow thanks for all that help...its supposed to be an identity though? sigh....perhaps its teachers mistake

17. myininaya Group Title

it is not

18. pottersheep Group Title

oh wait a minute if I change both sides then I can get them to be equal

19. myininaya Group Title

try x=pi/3 both sides have different outputs

20. myininaya Group Title

one side is approximately 1.366 the other side is .73205

21. Mertsj Group Title

No. It is not. You have shown that they indeed are reciprocals which was apparent by a casual inspection of the original problem.

22. myininaya Group Title

the equation does hold for some values but not all which makes it not an identity

23. pottersheep Group Title

Thanks you guys =) i guess i need much more practice

24. bluebrandon Group Title

when I simplified both sides after making them in terms of sin and cos I got (sinx-cosx)(sinx^2-cosx^2) ------------------------ cosx^2*sinx^2 for both the left hand side and the right hand side

25. bluebrandon Group Title

26. Mertsj Group Title

So how did you get the denominator to be a product instead of a difference?

27. myininaya Group Title

maybe he combine the fractions on top and on bottom but cos(x)sin(x) in the denominator of the bottom fraction would cancel with cos(x)sin(x) with the top's denominator

28. myininaya Group Title

so you wouldn't have that product anymore

29. pottersheep Group Title

yeah

30. pottersheep Group Title

omg! this is the second day we started trig functions and our teacher barely taught us anything......i wish she would start off with simpler question....

31. pottersheep Group Title

i still have like 18 questions to go...sigh...

32. pottersheep Group Title

trig identities :P not functions

33. myininaya Group Title

anyways potter we have already shown that this is not an identity do you have any questions as to why we concluded that?

34. Mertsj Group Title

Also, potter, are you absolutely sure that you copied the problem correctly?

35. pottersheep Group Title

Nope I understand...thank you everyone :) now i need to practice it

36. myininaya Group Title

ok have fun

37. pottersheep Group Title

lol you guys are calling me potter hehe. and yeah but our teacher does have a bad reputation of making typos lol

38. myininaya Group Title

is potter okay?

39. pottersheep Group Title

yep potter's fine :)

40. myininaya Group Title

harry potter? lol is that what you were thinking?

41. pottersheep Group Title

yeah xD

42. myininaya Group Title

it wouldn't be so bad to be the most powerful wizard of all time

43. pottersheep Group Title

yeah :( then i woulnt have to worry about these math problems lol

44. myininaya Group Title

you could just turn a math problem you didn't feel like working on into a bunny

45. pottersheep Group Title

LOL hahaha

46. pottersheep Group Title

too bad i cant give u another medal for that :P