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Math Help - stuck with 4 questions

  1. #1
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    stuck with 4 questions

    hi there, im currently stuck on 4 questions at the moment

    1. 9x^2 -14x + 1 = 0

    PROVE THE FOLLOWING IDENTITIES

    1. Sin2θSecθSinθTanθ

    2. tan2 A= sin2 A / 1 - sin2 A

    3. 2 - tan2 A ≡ 2Sec2 A - 3tan2 A

    hope someone out there can help would be much appreciated
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  2. #2
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    Quote Originally Posted by chaos View Post
    hi there, im currently stuck on 4 questions at the moment

    1. 9x^2 -14x + 1 = 0

    [snip]
    Use the quadratic formula. Are you familiar with it?
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Use the quadratic formula. Are you familiar with it?
    no im not unfortunatly, can you explain step by step please?
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  4. #4
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    Hello, chaos!

    1)\;\;9x^2 -14x + 1 \:=\: 0
    You don't know the Quadratic Formula ??

    x \;=\;\frac{\text{-}(\text{-}14) \pm\sqrt{(\text{-}14)^2 - 4(9)(1)}}{2(9)} \;=\;\frac{14 \pm\sqrt{160}}{18} \;=\;\frac{14 \pm4\sqrt{10}}{18} \;=\;\frac{7\pm2\sqrt{10}}{9}



    Prove the following identity: . 1)\;\;\sin^2\!\theta\sec\theta \:=\:\sin\theta\tan\theta
    The left side is: . \sin^2\!\theta\sec\theta \:=\:\sin^2\!\theta\cdot\frac{1}{\cos\theta} \;=\;\sin\theta\cdot\frac{\sin\theta}{\cos\theta} \;=\;\sin\theta\tan\theta



    2)\;\;\tan^2\!A\:=\:\frac{\sin^2\!A}{1 - \sin^2\!A}

    \text{The right side is: }\;\frac{\sin^2\!A}{\underbrace{1-\sin^2\!A}_{\text{This is }\cos^2\!A}} \;=\;\frac{\sin^2\!A}{\cos^2\!A} \;=\;\left(\frac{\sin A}{\cos A}\right)^2 \;=\;\tan^2\!A




    3)\;\;2 - \tan^2\!A \:=\:2\sec^2\!A - 3\tan^2\!A
    We should know that: . \sec^2\!A \:=\:\tan^2\!A + 1


    The right side is: . 2\sec^2\!A - 3\tan^2\!A \;\;=\;\;2(\tan^2\!A + 1) - 3\tan^2\!A

    . . . . . . . . . . . = \;\;2\tan^2\!A + 2 - 3\tan^2\!A \;\;=\;\;2 - \tan^2\!A

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  5. #5
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    thanks alot for the help guys, much appreciated
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