## LaddiusMaximus 5 years ago (sinx/(1+cosx)) +((1+cosx)/sinx) totally lost

1. anonymous

What do you have to do? prove it?

yeah=2cscx

3. myininaya

combine the fractions by finding a common denominator

yeah i got that, but what do I use, sinx? or 1+cosx?

5. anonymous

I'm working on it

6. anonymous

The Common Denominator will be sinx(1+cosx)

7. myininaya

what do you mean? how to combine these fractions: 2/3+5/7

so the common denominator is (sinx)1+cosx?

9. myininaya

sinx(1+cosx)

10. anonymous

yes that is the common denom

okay so now I have (sinx^2x(1+cosx)/sinx(1+cosx)) + (sinx(2+cos^2x)/sinx(1+cosx))?

12. myininaya

you will have (sinx)^2+(1+cosx)^2 in the numerator and sinx(1+cosx) is the denominator

sin^2x on both sides?

14. anonymous

I think you should have this so far: $\sin^2x + (1+cosx)^2/sinx(1+cosx)$

15. myininaya

ok look at your first fraction. what does the bottom lack so that it has the same denominator as the 2nd. sinx, right?

16. myininaya

multiply both the top and bottom by sinx on the first fraction

18. myininaya

now look at the 2nd fraction what does the bottom lack so that it has the same denominator as the first fraction, (1+cosx), right?

19. myininaya

so multiply the top and bottom of the 2nd fraction by (1+cosx). Now the fractions have the same denominator, you can write it as one fraction now

21. myininaya

so we have [sinx*sinx+(1+cosx)(1+cosx)]/[sinx(1+cosx)]

22. myininaya

[(sinx)^2+1+2cosx+(cosx)^2]/[sinx(1+cosx)]

23. anonymous

I think I got the complete answer/proof

24. myininaya

what is (sinx)^2+(cosx)^2

25. anonymous

=1

26. myininaya

lol gj, but i was laddius

27. anonymous

oops sorry

28. myininaya

29. myininaya

so the numerator becomes sin^2x + (1+cosx)^2?

31. anonymous

yes

32. myininaya

now foil that (1+cosx)^2

33. anonymous

exactly lol

so now its 1+2cosx+cosx^2?

35. anonymous

yes!

36. myininaya

+(sinx)^2

37. myininaya

now do you see (sinx)^2+(cosx)^2 anywhere in that numerator

ahhhhh pythagorean

39. myininaya

what do you have now?

so now I have 2+2cosx/sinx(1+cosx)

41. myininaya

ok what does the numerator terms have in common

2?

43. anonymous

HAHA ignore my last 3 posts..

44. myininaya

does anything cancel?

45. myininaya

jp?

the 2s cancel, right?

47. myininaya

no the 1+cosx do

48. myininaya

so you have 2/sinx

49. myininaya

which is ....

I have 2+2cosx in the numerator

cosx+cosx= 2cosx, right?

52. myininaya

didnt you say the numerator terms had a 2 in common so factor that out in you have 2(1+cosx)/[sinx(1+cosx)]

crap then the (1+cosx) cancel out and im left with 2cscx

54. myininaya

BINGO!