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anonymous

  • one year ago

how to verify frac(sec theta/ csc theta- cot theta -frac(sectheta/csc theta +cot theta = 2csc theta

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  1. Nnesha
    • one year ago
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    \[\large\rm \frac{ \sec \theta }{ \csc \theta -\cot \theta } -\frac{ \sec \theta }{ \csc \theta +\cot \theta }=2\csc \theta \] like this ?

  2. anonymous
    • one year ago
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    Yes

  3. anonymous
    • one year ago
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    Can you please help

  4. Nnesha
    • one year ago
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    ye gimme a sec let me do it i'll try to find easy way

  5. anonymous
    • one year ago
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    thanks

  6. Nnesha
    • one year ago
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    okay alright so first we should write sec and csc in terms of sin or cos csc = ?? sec equal what ? do you know ?

  7. anonymous
    • one year ago
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    1/sin - csc and sec =1/cos

  8. anonymous
    • one year ago
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    i meant csc=1/sin

  9. Nnesha
    • one year ago
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    yes right first deal with the denominator\[\large\rm \frac{ \sec \theta }{ \color{ReD}{\csc \theta -\cot \theta }} -\frac{ \sec \theta }{\color{reD}{ \csc \theta +\cot \theta }}=2\csc \theta \] \[\huge\rm \frac{ \sec \theta }{ \frac{ 1 }{ \sin }-\frac{ \cos }{ \sin}} - \frac{ \sec \theta }{ \frac{ 1 }{ \csc }+\frac{ \cos }{ \sin } }\] cot =cos over sin

  10. anonymous
    • one year ago
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    Yes i got that far then i get stuck

  11. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \sec \theta }{ \color{red}{\frac{ 1 }{ \sin }-\frac{ \cos }{ \sin}}} - \frac{ \sec \theta }{\color{red}{ \frac{ 1 }{ \csc }+\frac{ \cos }{ \sin } }}\] find common denominator of red part

  12. anonymous
    • one year ago
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    but i get 1/sin +cos/sin

  13. anonymous
    • one year ago
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    1-cos/sin - 1+cos/sin

  14. Nnesha
    • one year ago
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    if you have some work show it to me so i can find ur mistakes instead stating ovr

  15. anonymous
    • one year ago
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    1-cos/ sin -1+cos/sin sin/sin-sin/sin which equals 1 then i flip it to multiuply with 1/cos-1/cos

  16. Nnesha
    • one year ago
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    ugh gawd

  17. Nnesha
    • one year ago
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    brb i should refresh the page its lagging

  18. anonymous
    • one year ago
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    here is where i get lost

  19. Nnesha
    • one year ago
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    \[\large\rm \frac{ \sec }{ \frac{ 1+\cos }{ \sin } }-\frac{ \sec }{ \frac{ 1+\cos }{ \sin } }\]now multiply top with the reciprocal of the bottom fraction (change division to multiplication ) like for first one \[\large\rm \sec \times \frac{ \sin }{ 1-\cos } - \sec \times \frac{ \sin }{ 1+\cos }\]

  20. anonymous
    • one year ago
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    so then I get 1-cos(sec) /sin - 1+cos(sec)/sin

  21. Nnesha
    • one year ago
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    yes right we can change sec to 1 over cos \[\frac{ 1 }{ \cos } \times \frac{ \sin }{ 1-\cos } - \frac{ 1 }{ \cos } \times \frac{\sin }{ 1+\cos }\] \[\frac{ \sin }{ \cos(1-\cos) } -\frac{ \sin }{ \cos (1+\cos) }\] now find the common denominator

  22. anonymous
    • one year ago
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    cos (1-cos)(1+cos)

  23. Nnesha
    • one year ago
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    yes right \[\huge\rm \frac{ \sin\color{reD}{(1+\cos )}-\sin\color{reD}{(1-\cos)} }{ \cos (1-\cos)(1+\cos) }\] multiply sin with the denominator of 2nd fraction and multiply numerator of 2nd fraction with the denominator of 1st fraction that's how i got the red part now distribute parentheses by sin at the top and foil(1-cos)(1+cos)

  24. anonymous
    • one year ago
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    sin+sin cos - sin-sin cos

  25. anonymous
    • one year ago
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    then distribute? sorry wrong

  26. Nnesha
    • one year ago
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    sign mistake -sin times -cos = ?

  27. Nnesha
    • one year ago
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    yes foil (1-cos)(1+cos) which is at the denominator

  28. anonymous
    • one year ago
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    sin+sin cos

  29. anonymous
    • one year ago
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    cos (cos^2)

  30. Nnesha
    • one year ago
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    yes right \[\huge\rm \frac{ \sin +sincos-\sin +sincos }{ \cos(1-\cos)(1+\cos) }\] now (1-cos)(1+cos ) = ?

  31. Nnesha
    • one year ago
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    now it's not just cos ^2 |dw:1439658869096:dw| what do you get when you multiply first term by 1st term of 2nd parentheses

  32. anonymous
    • one year ago
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    Sorry its 1-cos^2

  33. Nnesha
    • one year ago
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    yes right yes right \[\huge\rm \frac{ \color{red}{\sin }+sincos\color{reD}{-\sin} +sincos }{ \cos(1-cos^2) }\] combine like terms (numerator) cos ^2 = ? do you remember the identity \[\sin^2 \theta + \cos^2 \theta=1\] solve this for cos^2

  34. anonymous
    • one year ago
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    2sin/1+cos^2

  35. Nnesha
    • one year ago
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    hmm no sin -sin = ?? sincos + sincos= ?? combine like terms

  36. anonymous
    • one year ago
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    sin-sin =0 and sincos+sincos=2sincos

  37. Nnesha
    • one year ago
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    yes right \[\huge\rm \frac{ 2sincos }{ \cos(1+\cos^2) }\] cos^2=what ? ^identity

  38. anonymous
    • one year ago
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    pythagorean identity soi have now 2sincos/cos(sin^2)

  39. Nnesha
    • one year ago
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    \[\huge\rm \sin^2 +\cos^2 =1\] solve for cos^2 subtract sin^2 both sides what od you get ?

  40. Nnesha
    • one year ago
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    do*

  41. anonymous
    • one year ago
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    1*sin^2

  42. anonymous
    • one year ago
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    1+sin^2

  43. Nnesha
    • one year ago
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    no subtract sin^2 both sides so should be 1-sin^2

  44. anonymous
    • one year ago
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    no 1-sin^2

  45. Nnesha
    • one year ago
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    yes right \[\huge\rm \frac{ 2sincos }{ \cos(1-cos^2) }\] replace cos^2 by 1-sin^2\[\huge\rm \frac{ 2sincos }{ \cos(1\color{Red}{-}(1-\sin^2))}\] distribute (1-sin^) by negative sign \[\huge\rm \frac{ 2sincos }{ cos(1-1+\sin^2) }\] 1-1=0 so \[\huge\rm \frac{ 2sincos }{ \cos \times \sin^2 }\]now divide

  46. anonymous
    • one year ago
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    2sin/sin^2

  47. anonymous
    • one year ago
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    no sorry 2/sin

  48. Nnesha
    • one year ago
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    yes \[ \frac{ 2 }{ \sin} = 2 \times \frac{ 1 }{ \sin }= ?\]

  49. anonymous
    • one year ago
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    2csc thanks

  50. Nnesha
    • one year ago
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    alright you can solve it in 5 minutes i found it easy method but there is another way to verify i really want to show that one too

  51. Nnesha
    • one year ago
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    @Loser66 please show ur work :P that was very easy

  52. Loser66
    • one year ago
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    You do it by yourself. You know it, right?

  53. Nnesha
    • one year ago
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    no..... ;P

  54. Loser66
    • one year ago
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    \(\dfrac{sec}{csc -cot}-\dfrac{sec}{csc +cot}=sec(\dfrac{1}{csc-cot}-\dfrac{1}{csc+cot})\) \(sec(\dfrac{csc+cot}{(csc-cot)(csc+cot)}-\dfrac{csc-cot}{(csc-cot)(csc+cot})\) =\(sec(\dfrac{csc+cot-csc+cot}{csc^2-cot^2})\) And we know that \(csc^2-cot^2=1\) hence it is \(sec(2cot)= 2\dfrac{1}{\cancel{cos}}\dfrac{\cancel{cos}}{sin}= 2csc\)

  55. Nnesha
    • one year ago
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    thanks.

  56. Loser66
    • one year ago
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    np

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