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Math Help - Prove this trig identity.

  1. #1
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    Prove this trig identity.

    Can someone solve it step by step. I tried two ways to solve it but didn't work.
    Attached Thumbnails Attached Thumbnails Prove this trig identity.-mathq.jpg  
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  2. #2
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    Its not clear to me what your question is, what are you supposed to show?

    It is clear that your steps are not correct. Can you state the question exactly how it is given to you?
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  3. #3
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    \frac{\sec^2{x}}{2 - \sec^2{x}} =

    \frac{1 + \tan^2{x}}{2 - (1 + \tan^2{x})} =

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

    multiply by \frac{\cos^2{x}}{\cos^2{x}} ...

    \frac{\cos^2{x} + \sin^2{x}}{\cos^2{x} - \sin^2{x}} =

    \frac{1}{\cos(2x)} =

    \sec(2x)
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  4. #4
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    Hello, mwok!

    Is this it? . \sec2\theta \:=\:\frac{\sec^2\!\theta}{2-\sec^2\!\theta}


    The left side is: . \sec2\theta \;=\;\frac{1}{\cos2\theta} \;=\;\frac{1}{2\cos^2\!\theta - 1}

    Divide top and bottom by \cos^2\!\theta\!:\quad \frac{\dfrac{1}{\cos^2\!\theta}}{\dfrac{2\cos^2\!\  theta}{\cos^2\!\theta} - \dfrac{1}{\cos^2\!\theta}} \;=\;\frac{\sec^2\!\theta}{2 - \sec^2\!\theta}

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  5. #5
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    Thanks!
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  6. #6
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    Quote Originally Posted by skeeter View Post
    \frac{\sec^2{x}}{2 - \sec^2{x}} =

    \frac{1 + \tan^2{x}}{2 - (1 + \tan^2{x})} =

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

    multiply by \frac{\cos^2{x}}{\cos^2{x}} ...

    \frac{\cos^2{x} + \sin^2{x}}{\cos^2{x} - \sin^2{x}} =

    \frac{1}{\cos(2x)} =

    \sec(2x)
    Why do you multiply with the cos/sin?
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  7. #7
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    Quote Originally Posted by mwok View Post
    Why do you multiply with the cos/sin?
    I didn't ... I multiplied by \frac{\cos^2{x}}{\cos^2{x}}
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