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AaronAndyson

  • one year ago

Trigo help..

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  1. AaronAndyson
    • one year ago
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    \[\frac{ \tan (A) }{ 1 - \cot(A) } + \frac{ \cot (A) }{ 1 - \tan(A) } = \sec(A)cosec(A) + 1\]

  2. AaronAndyson
    • one year ago
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    @Michele_Laino

  3. AaronAndyson
    • one year ago
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    @welshfella

  4. welshfella
    • one year ago
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    this one is tricky - i tried converting everything to sine and cosines but it only became more complicated.

  5. welshfella
    • one year ago
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    it miggt help to convert all the cot A to tan A

  6. welshfella
    • one year ago
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    - that doesn't help

  7. AaronAndyson
    • one year ago
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    @mathslover

  8. AaronAndyson
    • one year ago
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    @paki @IrishBoy123 @welshfella

  9. Michele_Laino
    • one year ago
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    please try this substitution: \[\cot A = \frac{1}{{\tan A}}\]

  10. welshfella
    • one year ago
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    thats what I tried but didnt get very far - I'll have another try

  11. AaronAndyson
    • one year ago
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    It's getting complicated >.<

  12. AaronAndyson
    • one year ago
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    @Michele_Laino

  13. welshfella
    • one year ago
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    i boiled it down to t^3 - 1 ------ t(t - 1) where t = tan A

  14. imqwerty
    • one year ago
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    (Tan A/1-Cot A) +( Cot A / 1 - Tan A )= Sec A.Cosec A + 1 = (sin A/cos A) / [1- (cos A/sin A)] + (cos A/sin A) / [1- (sin A/cos A)] = (sin A/cos A) / [(sin A - cos A)/sin A] + (cos A/sin A) / [(cos A - sin A) /cos A] = (cos^2 A) / [sin A(cos A - sin A)] - (sin^2 A) / [cos A(cos A - sin A)] = (cos^3 A - sin^3 A) / [(cos A * sin A) (cos A - sin A)] = (1/cos A * sin A) (sin^2 A + cos^2 A + sin A cos A) = (1/cos A * sin A) (1+ sin A cos A) = [1 / (cos A sin A) + (cos A sin A) / (cos A sin A)] =1+cscAsecA

  15. Michele_Laino
    • one year ago
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    for example the first fraction at the left side will become: \[\Large \frac{{\tan A}}{{1 - \cot A}} = \frac{{\tan A}}{{1 - \frac{1}{{\tan A}}}} = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}}\]

  16. AaronAndyson
    • one year ago
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    second one (cotA)^2/cotA - 1

  17. AaronAndyson
    • one year ago
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    right?

  18. Michele_Laino
    • one year ago
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    the second fraction will become: \[\Large \begin{gathered} \frac{{\cot A}}{{1 - \tan A}} = \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}} = \frac{1}{{\tan A\left( {1 - \tan A} \right)}} = \hfill \\ = - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\ \end{gathered} \]

  19. AaronAndyson
    • one year ago
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    Okay! What next?

  20. Michele_Laino
    • one year ago
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    we have to add those fractions together, like below: \[\Large \begin{gathered} \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\ \end{gathered} \]

  21. welshfella
    • one year ago
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    yep that's what i got

  22. Michele_Laino
    • one year ago
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    next we have this factorization: \[\large {\left( {\tan A} \right)^3} - 1 = \left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}\]

  23. AaronAndyson
    • one year ago
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    you have subtracted the 2 fraction ... why?

  24. AaronAndyson
    • one year ago
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    http://prntscr.com/7x8oyo shouldnt the sign by +?

  25. AaronAndyson
    • one year ago
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    @Michele_Laino

  26. Michele_Laino
    • one year ago
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    I have not subtracted the 2 fractions, since the second fraction can be rewritten as follows: \[\Large \begin{gathered} \frac{{\cot A}}{{1 - \tan A}} = \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}} = \frac{1}{{\tan A\left( {1 - \tan A} \right)}} = \hfill \\ = - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\ \end{gathered} \]

  27. AaronAndyson
    • one year ago
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    ok! what next?

  28. Michele_Laino
    • one year ago
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    using the factorization above, I can write: \[\large \begin{gathered} \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} \hfill \\ \hfill \\ \end{gathered} \]

  29. AaronAndyson
    • one year ago
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    ok!

  30. Michele_Laino
    • one year ago
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    next we have: \[\large \begin{gathered} \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\ \end{gathered} \]

  31. AaronAndyson
    • one year ago
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    i did not get the last time up there

  32. Michele_Laino
    • one year ago
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    now, we can simplify as follows: \[\Large \begin{gathered} \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} + 1}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}}}\frac{{\cos A}}{{\sin A}} = \hfill \\ \hfill \\ = \frac{1}{{\sin A\cos A}} = \sec A\csc A \hfill \\ \end{gathered} \]

  33. Michele_Laino
    • one year ago
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    and then: \[\Large \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 = \sec A\csc A + 1\]

  34. AaronAndyson
    • one year ago
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    I didnt get a thing ......

  35. Michele_Laino
    • one year ago
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    ok! I redo my procedure

  36. Michele_Laino
    • one year ago
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    if we make this substitution: \[\cot A = \frac{1}{{\tan A}}\]

  37. Michele_Laino
    • one year ago
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    we can write this: \[\frac{{\tan A}}{{1 - \cot A}} = \frac{{\tan A}}{{1 - \frac{1}{{\tan A}}}} = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}}\]

  38. Michele_Laino
    • one year ago
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    and: \[\begin{gathered} \frac{{\cot A}}{{1 - \tan A}} = \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}} = \frac{1}{{\tan A\left( {1 - \tan A} \right)}} = \hfill \\ = - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\ \end{gathered} \]

  39. Michele_Laino
    • one year ago
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    am I right?

  40. AaronAndyson
    • one year ago
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    yes.

  41. Michele_Laino
    • one year ago
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    now I develop your original expression as follows: \[\begin{gathered} \frac{{\tan A}}{{1 - \cot A}} + \frac{{\cot A}}{{1 - \tan A}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \end{gathered} \]

  42. AaronAndyson
    • one year ago
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    ok?

  43. Michele_Laino
    • one year ago
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    are you sure?

  44. AaronAndyson
    • one year ago
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    yes..

  45. Michele_Laino
    • one year ago
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    ok! Now the common denominator is: \[{\tan A\left( {\tan A - 1} \right)}\]

  46. AaronAndyson
    • one year ago
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    yes

  47. Michele_Laino
    • one year ago
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    so, I can write this: \[\begin{gathered} \frac{{\tan A}}{{1 - \cot A}} + \frac{{\cot A}}{{1 - \tan A}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ \end{gathered} \]

  48. Michele_Laino
    • one year ago
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    where, at the last step I have used my factorization: \[{\left( {\tan A} \right)^3} - 1 = \left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}\]

  49. AaronAndyson
    • one year ago
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    ok

  50. Michele_Laino
    • one year ago
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    now I can cancel out tanA-1 from the numerator and denominator of the last fraction

  51. AaronAndyson
    • one year ago
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    yup

  52. Michele_Laino
    • one year ago
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    so I get: \[\begin{gathered} \frac{{\tan A}}{{1 - \cot A}} + \frac{{\cot A}}{{1 - \tan A}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\ \end{gathered} \]

  53. AaronAndyson
    • one year ago
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    the last step ..me no get...

  54. Michele_Laino
    • one year ago
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    here is the last step: \[\begin{gathered} \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1 + \tan A}}{{\tan A}} = \hfill \\ \hfill \\ = \frac{{\left\{ {{{\left( {\tan A} \right)}^2} + 1} \right\} + \tan A}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\ \end{gathered} \]

  55. AaronAndyson
    • one year ago
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    shouldnt both the tan be cancelled?

  56. Michele_Laino
    • one year ago
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    not exactly, look at this: \[\begin{gathered} \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1 + \tan A}}{{\tan A}} = \hfill \\ \hfill \\ = \frac{{\left\{ {{{\left( {\tan A} \right)}^2} + 1} \right\} + \tan A}}{{\tan A}} = \frac{{\left\{ {{{\left( {\tan A} \right)}^2} + 1} \right\}}}{{\tan A}} + \frac{{\tan A}}{{\tan A}} \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\ \end{gathered} \]

  57. AaronAndyson
    • one year ago
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    okay!

  58. Michele_Laino
    • one year ago
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    ok! Now we use the substitution: \[\tan A = \frac{{\sin A}}{{\cos A}}\]

  59. AaronAndyson
    • one year ago
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    ok

  60. Michele_Laino
    • one year ago
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    so we can write this: \[\begin{gathered} \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} + 1}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} = \frac{1}{{\sin A\cos A}} \hfill \\ \end{gathered} \]

  61. Michele_Laino
    • one year ago
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    \[\large \begin{gathered} \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} + 1}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} = \frac{1}{{\sin A\cos A}} \hfill \\ \end{gathered} \]

  62. AaronAndyson
    • one year ago
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    THANKS!

  63. Michele_Laino
    • one year ago
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    :)

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