## abichu Group Title 6) If sec theta+tan theta = P. PT sin theta=P^2-1/P^2+1 3 years ago 3 years ago

1. myininaya

we want to show the second part?

2. abichu

yea!

3. myininaya

what is T?

4. abichu

PT= PROVE THAT

5. myininaya

so we get to suppose $\sec(\theta)+\tan(\theta)=P$ and we want to show $PT \sin(\theta)=\frac{P^2-1}{P^2+1}$?

6. myininaya

oh ok

7. abichu

8. myininaya

$\sec(\theta)+\tan(\theta)=\frac{1}{\cos(\theta)}+\frac{\sin(\theta)}{\cos(\theta)}=\frac{1+\sin(\theta)}{\cos(\theta)}=P$ thinking....

9. myininaya

$\frac{1+\sin(\theta)}{\cos(\theta)}*\frac{1-\sin(\theta)}{1-\sin(\theta)}=\frac{1-\sin^2(\theta)}{\cos(\theta)(1-\sin(\theta))}$ $=\frac{\cos^2(\theta)}{\cos(\theta)(1-\sin(\theta))}=\frac{\cos(\theta)}{1-\sin(\theta)}=P$ hmmm....

10. abichu

solved?

11. myininaya

no lol

12. myininaya

i havent solved it yet

13. myininaya

maybe i should try this on paper first

14. abichu

what bout my prev.sums? can you solve it?

15. myininaya

let me try this one first

16. abichu

sure! :)

17. myininaya

if $\sec(\theta)+\tan(\theta)=P$ then $(\sec(\theta)+\tan(\theta))(\sec(\theta)-\tan(\theta))=P(\sec(\theta)-\tan(\theta))$ but $(\sec(\theta)+\tan(\theta))(\sec(\theta)-\tan(\theta))=\sec^2(\theta)-\tan^2(\theta)=1$ so we have $1=P(\sec(\theta)-\tan(\theta))$ => $\sec(\theta)-\tan(\theta)=\frac{1}{P}$ lets see if we can use this

18. myininaya

lets see we have that $\sec(\theta)+\tan(\theta)=\frac{1+\sin(\theta)}{\cos(\theta)}$ and $\sec(\theta)-\tan(\theta)=\frac{1-\sin(\theta)}{\cos(\theta)}$ adding these together we get $\frac{1+\sin(\theta)}{\cos(\theta)}+\frac{1-\sin(\theta)}{\cos(\theta)}=P+\frac{1}{P}$ $\frac{2}{\cos(\theta)}=\frac{P^2+1}{P}$ $\frac{2P}{P^2+1}=\cos(\theta)$

19. myininaya

but if we imagine we have a right triangle and we draw theta somewhere then cos tells adj/hyp=2p/(P^2+1) we can find the opposite of side of theta opp^2+adj^2=hyp^2 $opp^2=hyp^2-adj^2=(P^2+1)^2-(2P)^2=P^4+2P^2+1-2^2P^2$ so we have $opp=\sqrt{P^4-2P^2+1}$ but sin=opp/hyp

20. myininaya

$\sin(\theta)=\frac{\sqrt{P^4-2P^2+1}}{P^2+1}=\frac{\sqrt{(P^2-1)^2}}{P^2+1}$

21. myininaya

$\sin(\theta)=\frac{P^2-1}{P^2+1}$

22. myininaya

any questions?

23. abichu

yea..can you solve the other problems?

24. myininaya

zarkon?

25. abichu

what?

26. myininaya

i was looking for someone

27. Zarkon

yes

28. abichu

and can you temme one thing..can you temme the stepin correct order so that i can copy it in my rough boook! ;)

29. myininaya

hey zarkon i had to think on it isn't that crazy?

30. Zarkon

yes

31. myininaya

the problem

32. Zarkon

trying to think if there is an easier way

33. myininaya

i have no clue it was hard for me to see that way

34. myininaya

what is temme?

35. myininaya

i don't know that word

36. myininaya

37. myininaya

well i mean i guess its small but i don't know it

38. myininaya

tell me?

39. myininaya

temme=tell me?

40. myininaya

i thought i showed u the steps zarkon thinks there may be an easier way though

41. myininaya

zarkon i did this for you :)

42. myininaya

lol

43. Zarkon

;)

44. Zarkon

45. myininaya

and for abichu dont feel left out

46. Zarkon

get to $\frac{1+sin(\theta)}{\cos(\theta)}=p$ ...

47. abichu

i am soo weak in math..that's why am seeking help from you guys! if you dont mind help me! ;)

48. myininaya

we are helping

49. Zarkon

$1+\sin(\theta)=p\cos(\theta)$

50. myininaya

i showed you one way and zarkon is gonna show you another it appears

51. Zarkon

$1+\sin(\theta)=p\sqrt{1-\sin^2(\theta)}$

52. Zarkon

square both sides...

53. myininaya

omg i see where you are going

54. Zarkon

$1+2\sin(\theta)+\sin^2(\theta)=p^2(1-\sin^2(\theta))$

55. myininaya

you going to write is as a quadratic

56. myininaya

and solve for sin(theta)

57. abichu

can you tel me from the 1st step if you dot mind?!

58. Zarkon

move stuff to one side $1-p^2+2\sin(\theta)+(1+p^2)\sin^2(\theta)=0$

59. myininaya

brillant!

60. Zarkon

$1-p^2+2x+(1+p^2)x^2=0$

61. Zarkon

62. Zarkon

fun stuff :)

63. myininaya

yes

64. myininaya

you are a showoff lol

65. Zarkon

lol...now abichu has two ways to do it

66. abichu

i donno how to solve it quadratic..can you doo it n give?

67. myininaya

$x=\frac{-2 \pm \sqrt{(2)^2-4(1+p^2)(1-p^2)}}{2(1+p^2)}$