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Math Help - Please help with this Trig identity!!!!!

  1. #1
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    Exclamation Please help with this Trig identity!!!!!

    For some reason I CANT get the answer for


    (1-sin^2X)/(sinX-cscX)= -sinX

    PLEASE PLEASE PLEASE!
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  2. #2
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    Quote Originally Posted by amanda0603 View Post
    For some reason I CANT get the answer for


    (1-sin^2X)/(sinX-cscX)= -sinX

    PLEASE PLEASE PLEASE!
    sin(x) - csc(x) = sin(x) - \frac{1}{sin(x)}

    set the same denominator:

    \frac{sin^2(x) -1}{sin(x)}

    1-sin^2(x) \times \frac{sin(x)}{sin^2(x) -1}

    Note that 1-sin^2(x) = -(sin^2(x) - 1)

    -(sin^2(x)-1) \times \frac{sin(x)}{sin^2(x) -1}

    which cancels to -sin(x)
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  3. #3
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    well, sin^2(x)+cos^2(x)=1, so 1-sin^2(x)= cos^2(x)
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  4. #4
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    Hello, Amanda!

    Another approach . . .


    \frac{1-\sin^2\!x}{\sin x-\csc x} \:=\: -\sin x

    \text{We have: }\;\frac{1-\sin^2\!x}{\sin x-\csc x} \:=\:\frac{1-\sin^2\!x}{\sin x - \frac{1}{\sin x}}

    Multiply by \frac{\sin x}{\sin x}\!:\;\;\frac{\sin x}{\sin x}\cdot\frac{1-\sin^2\!x}{\sin x - \frac{1}{\sin x}} \;=\;\frac{\sin x(1 - \sin^2\!x)}{\sin^2\!x - 1}

    Factor and reduce: . \frac{-\sin x\,({\color{red}\rlap{////////}}\sin^2\!x-1)}{{\color{red}\rlap{////////}}\sin^2\!x-1} \;=\; -\sin x

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