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Math Help - trig identity help.

  1. #1
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    trig identity help.

    Write the following in their simplest form, involving only one trigonometric function:

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

     \frac{tan \theta}{ 1 + tan^{2} \theta - 2}

     \frac{tan\theta}{tan^2 \theta-1}

    I am stuck at this point, I know I have to get it to look like the  tan2\theta expansion,

    any hints/help appreciated.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Tweety View Post
     1 + tan^{2}\theta = sec^{2} \theta

     \frac{tan \theta}{ 1 + tan^{2} \theta - 2}

     \frac{tan\theta}{tan^2 \theta-1}

    I am stuck at this point, I know I have to get it to look like the  tan2\theta expansion,

    any hints/help appreciated.
    Hi Tweety,

    Maybe you could do something like this.

     \frac{tan\theta}{tan^2 \theta-1} = -\frac{\tan \theta}{1-\tan^2 \theta} \cdot \frac{2}{2}=-\frac{2\tan \theta}{2(1-\tan^2 \theta)}=-\frac{1}{2} \tan 2 \theta
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  3. #3
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    Quote Originally Posted by masters View Post
    Hi Tweety,

    Maybe you could do something like this.

     \frac{tan\theta}{tan^2 \theta-1} = -\frac{\tan \theta}{1-\tan^2 \theta} \cdot \frac{2}{2}=-\frac{2\tan \theta}{2(1-\tan^2 \theta)}=-\frac{1}{2} \tan 2 \theta
    thanks,

    Just one question. How did you get  1-tan^{2} \theta from  tan^{2} \theta - 1 ?

    Did you multiply top and bottom by minus one?
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  4. #4
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    Quote Originally Posted by Tweety View Post

    Did you multiply top and bottom by minus one?
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