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Thread: minimum positive value.

  1. #1
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    minimum positive value.

    If $\displaystyle a,b$ are positive quantities and $\displaystyle a\geq b$.Then find Minimum positive value of $\displaystyle asec\theta-btan\theta$
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  2. #2
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    $\displaystyle \displaystyle a \sec \theta \tan \theta - b \sec^{2} \theta = 0 $

    $\displaystyle \displaystyle \frac{a}{b} = \frac{\sec \theta}{\tan \theta} $

    $\displaystyle \displaystyle \frac{a}{b} = \frac{1}{\sin \theta} $

    $\displaystyle \displaystyle \frac{b}{a} = \sin \theta $

    $\displaystyle \displaystyle \theta = \arcsin \frac {b}{a} + 2 \pi n, \ n = 0, \ \pm 1, \pm2, ...$
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  3. #3
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    but ans is in terms of $\displaystyle a$ and $\displaystyle b$ i.e

    $\displaystyle asec\theta-btan\theta \geq \sqrt{a^2-b^2}$
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  4. #4
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    Hello, jacks!

    but ans is in terms of $\displaystyle \,a$ and $\displaystyle \,b$

    $\displaystyle a\sec\theta-b\tan\theta \:\geq\: \sqrt{a^2-b^2}$

    Random Variable did the hard work for you . . .


    $\displaystyle \text{We have: }\:f(\theta) \:=\:a\sec\theta - b\tan\theta$

    $\displaystyle \text{He found that }\,f'(\theta) = 0\,\text{ when }\,\sin\theta \,=\,\dfrac{b}{a}$

    . . $\displaystyle \text{Hence: }\;\sec\theta \,=\,\dfrac{a}{\sqrt{a^2-b^2}},\;\;\tan\theta \,=\,\dfrac{b}{\sqrt{a^2-b^2}}$


    Substitute into the function:

    . . $\displaystyle \displaystyle a\sec\theta - b\tan\theta \;\ge\;a\frac{a}{\sqrt{a^2-b^2}} - b\frac{b}{\sqrt{a^2-b^2}}$

    . . . . . . . . . . . . . .$\displaystyle \displaystyle =\;\frac{a^2}{\sqrt{a^2-b^2}} - \frac{b^2}{\sqrt{a^2-b^2}}$

    . . . . . . . . . . . . . .$\displaystyle \displaystyle =\;\frac{a^2-b^2}{\sqrt{a^2-b^2}} \;=\;\sqrt{a^2-b^2}$


    $\displaystyle \text{Therefore: }\;a\sec\theta - b\tan\theta \;\ge \;\sqrt{a^2+b^2}$

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    Thanks Random variable and Sorobon.
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