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Thread: Need Help Simplifying trig expression

  1. #1
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    Red face Need Help Simplifying trig expression

    Well first off I am not the best simplifier, yet i still am managable, but now with the whole trig function ordeal and it makes it twice as hard.


    Could someone please show me step by step on how to simplify
    [(tan q + sec q)^2 +1] / [2sec q]

    Thanks much!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Mr_Green View Post
    Well first off I am not the best simplifier, yet i still am managable, but now with the whole trig function ordeal and it makes it twice as hard.


    Could someone please show me step by step on how to simplify
    [(tan q + sec q)^2 +1] / [2sec q]

    Thanks much!
    First replace all functions by $\displaystyle \sin$s and $\displaystyle \cos$s:

    $\displaystyle
    f(\theta)=\frac{(\tan(\theta)+\sec(\theta))^2+1)}{ 2 \sec(\theta)}=
    $$\displaystyle
    \left(\left(\frac{s}{c}+\frac{1}{c}\right)^2+1\rig ht)\frac{c}{2}
    $,

    where $\displaystyle c$ and $\displaystyle s$ are shorthand for $\displaystyle \cos(\theta)$ and $\displaystyle \sin(\theta)$, so:

    $\displaystyle f(\theta)=
    \left(\left(\frac{s+1}{c}\right)^2+1\right)\frac{c }{2}=
    \left(\left(\frac{s^2+2s+1}{c^2}\right)+1\right)\f rac{c}{2}=$$\displaystyle
    \left(\frac{s^2+2s+1+c^2}{c^2}\right)\frac{c}{2}=
    \left(\frac{2s+2}{c^2}\right)\frac{c}{2}
    $

    .........$\displaystyle
    =\left(\frac{s+1}{c}\right)=\tan(\theta)+\sec(\the ta)
    $

    RonL
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  3. #3
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    Hello, Mr_Green!

    Simplify: .$\displaystyle \frac{(\tan\theta + \sec\theta)^2 +1}{2\sec\theta}$

    We have: .$\displaystyle \frac{\tan^2\theta + 2\tan\theta\sec\theta + \sec^2\theta + 1}{2\sec\theta}$ $\displaystyle = \;\frac{(\tan^2\theta + 1) + \sec^2\theta + 2\sec\theta\tan\theta}{2\sec\theta} $

    . . . . . .$\displaystyle = \;\frac{\sec^2\theta + \sec^2\theta + 2\sec\theta\tan\theta}{2\sec\theta}\;=\;\frac{2\se c^2\theta + 2\sec\theta\tan\theta}{2\sec\theta}$


    Factor and reduce: . $\displaystyle \frac{2\sec\theta(\sec\theta + \tan\theta)}{2\sec\theta} \;= \;\sec\theta + \tan\theta $

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