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Verify Identity

  1. Jun 21, 2005 #1
    Hello All

    I am having problems Verifying this identity

    (COS A) / (1-SIN A) = SEC A + TAN A

    I can get the RHS = (1 + SIN A)/COS A

    But this does not equal the LHS

    Any advice is welcome.

    THanks
     
  2. jcsd
  3. Jun 21, 2005 #2
    yes, you are right, now cross-multiply:

    [tex]cos^2 a = (1 + sin a) (1 - sin a)[/tex]
    [tex]cos^2 a = 1 - sin^2 a[/tex]
    [tex]sin^2 a + cos^2 a = 1[/tex]
     
  4. Jun 21, 2005 #3
    sec A + tan A = (1/cosA) + sinA/cosA = (cosA + sinA*cosA)/(cosA)^2 =

    cosA(1 + sinA)/(1-sin^2A) = cos A / (1 + sinA) :cool:
     
  5. Jun 21, 2005 #4
    does [tex]sin^2 a + cos^2 a = 1[/tex] verify the idenity?
     
  6. Jun 21, 2005 #5

    James R

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    Divide the LHS top and bottom by cos A:

    [tex]\frac{\cos A}{1-\sin A} = \frac{1}{\sec A - \tan A}[/tex]
    [tex]=\frac{1}{(\sec A - \tan A)}\frac{(\sec A + \tan A)}{(\sec A + \tan A)}[/tex]
    [tex]=\frac{\sec A + \tan A}{\sec^2 A - \tan^2 A}[/tex]

    But

    [tex]\sec^2 A - \tan^2 A = \frac{1 - \sin^2 A}{\cos^2 A} = 1[/tex]

    So, the result follows.
     
  7. Jun 22, 2005 #6

    HallsofIvy

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    Yes, it does: (1+ sin A)/cos A= 1/cos A+ sin A/cos A= sec A+ tan A.
     
  8. Jun 23, 2005 #7
    Thanks All Still kind of confused.

    HallsofIvy: doesn't that just change back the RHS that I did??
     
  9. Jun 23, 2005 #8

    James R

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    powp:

    Does this help?

    [tex]\frac{1 + \sin A}{\cos A} = \frac{(1 + \sin A)(1 - \sin A)}{\cos A(1 - \sin A)}[/tex]

    [tex]=\frac{1 - \sin^2 A}{\cos A (1 - \sin A)}[/tex]

    [tex]= \frac{\cos^2 A}{\cos A (1 - \sin A)}[/tex]

    [tex]= \frac{\cos A}{1- \sin A}[/tex]
     
  10. Jun 23, 2005 #9

    uart

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    Yes that is perhaps the most fundemental of all trig identities. It's actually just Pythagoras Thm for a RHT with unit length hypotemus.
     
  11. Jun 28, 2005 #10
    Is this correct??

    Hello All

    I need to find all solutions to the following did i do it correct?

    Sin 2x = 2Tan 2x

    2Sinx Cosx = 2(2Tanx / 1-Tan^2x)

    Sinx Cosx = (2(Sinx / Cosx)/(cos^2x-Sin^2x / Cos^2x))

    Sinx Cosx = 2(Sinx / Cosx) X (Cos^2x / cos^2x-Sin^2x)

    Cosx = 2 (Cosx / Cos^2x - Sin^2x)

    1 = (2 / Cos^2x - Sin^2x)

    Cos^2x - Sin^2x = 2

    1 - Sinx^2 - Sin^2x = 2

    2Sin^2x = -1
    Sin^2x = -1/2

    Sinx = -1/SQROOT(2)



    Is this correct??
     
  12. Jun 28, 2005 #11

    arildno

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    Dearly Missed

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