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Math Help - Inverse trigonometric function

  1. #1
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    Exclamation Inverse trigonometric function

    arc tan [1/{(x^2)-1}^(1/2)].....mod x >1.
    I am asked to write it in the simplest form.....how do i proceed?
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  2. #2
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    Question

    Quote Originally Posted by tariq_h_tauheed View Post
    arc tan [1/{(x^2)-1}^(1/2)].....mod x >1.
    I am asked to write it in the simplest form.....how do i proceed?
    How does your book define "simplest form" for similar exercises? Are you maybe supposed to "rationalize the denominator" for "simplest form"...?
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  3. #3
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    Exclamation

    The answer in the book is itself an inverse trigonometric function....actually the answer is in the order of arc sec......but i don't know the way to proceed.
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  4. #4
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    Hello, tariq_h_tauheed!

    Write in simplest form: . \arctan\left(\frac{1}{\sqrt{x^2-1}}\right)

    Let: . \theta \;=\;\arctan\left(\frac{1}{\sqrt{x^2-1}}\right) \quad\hdots\quad\text{then: }\:\tan\theta \:=\:\frac{1}{\sqrt{x^2-1}} \:=\:\frac{opp}{adj}


    \theta is in a right triangle with: opp = 1,\;adj = \sqrt{x^2-1}


    Code:
                            *
                         *  |
                 x    *     |
                   *        | 1
                *           |
             * θ            |
          * - - - - - - - - *
                √(x-1)
    Using Pythagorus, we get: . hyp = x

    Then: . \csc\theta \:=\:\frac{x}{1} \:=\:x


    Therefore: . \theta \;=\;\text{arccsc}\,x . .
    (simplest form )
    .
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