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Math Help - 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 \sins and \coss:

    <br />
f(\theta)=\frac{(\tan(\theta)+\sec(\theta))^2+1)}{  2 \sec(\theta)}=<br />
<br />
\left(\left(\frac{s}{c}+\frac{1}{c}\right)^2+1\rig  ht)\frac{c}{2}<br />
,

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

    f(\theta)=<br />
\left(\left(\frac{s+1}{c}\right)^2+1\right)\frac{c  }{2}=<br />
\left(\left(\frac{s^2+2s+1}{c^2}\right)+1\right)\f  rac{c}{2}= <br />
\left(\frac{s^2+2s+1+c^2}{c^2}\right)\frac{c}{2}=<br />
\left(\frac{2s+2}{c^2}\right)\frac{c}{2}<br />

    ......... <br />
=\left(\frac{s+1}{c}\right)=\tan(\theta)+\sec(\the  ta)<br />

    RonL
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  3. #3
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    Hello, Mr_Green!

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

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

    . . . . . . = \;\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: . \frac{2\sec\theta(\sec\theta + \tan\theta)}{2\sec\theta} \;= \;\sec\theta + \tan\theta

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