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  1. #1
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    Problem Help

    Having trouble with this problem....

    Establish the Identity:
    (sec^2t)-(csc^2t)=csc^2t(sec^2t-2)


    and this ones seems easy but im not sure...

    Find the exact Value:
    csc(tan^-1)(-.5)



    hummm,


    Thanks if anyone can help!
    Last edited by gabrie30; June 3rd 2008 at 08:52 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by gabrie30 View Post
    Having trouble with this problem....

    sec^2-csc^2=csc^2(sec^2-2) "after ^2 there is theta"



    and this ones seems easy but im not sure...

    csc(tan^-1(-.5)



    hummm,


    Thanks if anyone can help!
    what are you supposed to be doing here...?
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  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
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    Quote Originally Posted by Jhevon View Post
    what are you supposed to be doing here...?
    Just moo

    @ gabrie30 : can you put parenthesis ? and use t for theta
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    what are you supposed to be doing here...?


    Finding the exact value for the inverse tan and "establish the identity" for the other
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  5. #5
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    Quote Originally Posted by gabrie30 View Post
    Establish the Identity:
    (sec^2t)-(csc^2t)=csc^2t(sec^2t-2)
    LHS = \frac{1}{cos^2t} - \frac{1}{sin^2t}
    = \frac{sin^2t- cos^2t}{cos^2t\,sin^2t}
     = \frac{(1-cos^2t) - cos^2t}{cos^2t\,sin^2t}
    = \frac{1}{cos^2t\,sin^2t} - \frac{2\not cos^2t}{\not cos^2t\,sin^2t}
    = \frac{1}{sin^2t} \left(\frac{1}{cos^2t} - 2 \right)
    = csc^2t ( sec^2t -2)

    Find the exact Value:
    csc(tan^{-1}(-.5))

    Note that arctan(-.5) is an angle
    First let  y = arctan (-.5)
     tan y = -.5
    If you recall basic trigonometric ratios, tangent =  \frac{opposite\,side}{adjacent\,side} = - \frac{1}{2} in this case.
    Now we know that the opposite side is  \pm 1 when the adjacent side is \mp 2
    Using Pythagoras's theorem, we can deduce that the hypotenuse is  \sqrt{5}
     csc \,y = \frac{hypotenuse}{opposite}  = \pm \frac{\sqrt5}{1}
    = \pm \sqrt5

    You might want to draw a triangle if it helps you to visualise
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  6. #6
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    [quote=Gusbob;153603] LHS = \frac{1}{cos^2t} - \frac{1}{sin^2t}
    = \frac{sin^2t- cos^2t}{cos^2t\,sin^2t}
     = \frac{(1-cos^2t) - cos^2t}{cos^2t\,sin^2t}
    = \frac{1}{cos^2t\,sin^2t} - \frac{2\not cos^2t}{\not cos^2t\,sin^2t}


    Can someone make this step more clear.



    = \frac{1}{sin^2t} \left(\frac{1}{cos^2t} - 2 \right)

    = csc^2t ( sec^2t -2)
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