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Math Help - Somebody plz try out proving this indentity....

  1. #1
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    Somebody plz try out proving this indentity....

    If tanΘ = 4/3,


    then prove that:√(1-sin Θ)/(1+sin Θ) = 1/3
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  2. #2
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    There is difficulity with this problem as stated.
    If \tan(\theta)=\frac{4}{3} then \sin(\theta)=\pm\frac{4}{5}.
    That identity is true for only one of those values.
    Did you leave some information out?
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  3. #3
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    Quote Originally Posted by Plato View Post
    There is difficulity with this problem as stated.
    If \tan(\theta)=\frac{4}{3} then \sin(\theta)=\pm\frac{4}{5}.
    That identity is true for only one of those values.
    Did you leave some information out?
    umm....nope...
    i posted the question as it was given in my book...
    what do you think will be the correct identity then??
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  4. #4
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    Quote Originally Posted by snigdha View Post
    umm....nope...
    i posted the question as it was given in my book...
    what do you think will be the correct identity then??
    It is correct if \theta\in I. Otherwise no.
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  5. #5
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    Draw a 3-4-5 triangle where the hypotenuse is 5, the vertical segment is 4 and theta is opposite to it. Let theta = y. Therefore, siny is 4/5 and cosy is 3/5 Since tany = siny/cosy, tany = (4/5)/(3/5) or 4/3.
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  6. #6
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    Quote Originally Posted by blackcompe View Post
    Draw a 3-4-5 triangle where the hypotenuse is 5, the vertical segment is 4 and theta is opposite to it. Let theta = y. Therefore, siny is 4/5 and cosy is 3/5 Since tany = siny/cosy, tany = (4/5)/(3/5) or 4/3.

    But we are required to prove √(1-sin Θ)/(1+sin Θ) = 1/3
    tanΘ = 4/3 is the information given to us....
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  7. #7
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    Hint :

    \sin \theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\frac{1}{\sec^2\theta}} = \sqrt{1-\frac{1}{1+\tan^2\theta}}
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  8. #8
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    Quote Originally Posted by danielomalmsteen View Post
    Hint :

    \sin \theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\frac{1}{\sec^2\theta}} = \sqrt{1-\frac{1}{1+\tan^2\theta}}

    wow!!
    you are genius...!! how did you work this out..?!
    i had spent almost 3days on this problem in vain..!!!
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