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Thread: A simple problem...

  1. #1
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    A simple problem...

    If sec θ = x + 1/(4x), then prove that sec θ + tan θ = 2x or 1/(2x) .
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  2. #2
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    Hello, snigdha!

    $\displaystyle \text{If }\sec\theta \:=\: x + \frac{1}{4x},\,\text{ then prove that: }\:\sec\theta + \tan\theta \:=\: 2x\,\text{ or }\,\frac{1}{2x}$
    We have: .$\displaystyle \sec\theta \:=\:\frac{4x^2+1}{4x} \;=\;\frac{hyp}{adj}$

    $\displaystyle \theta$ is in a right triangle with: .$\displaystyle adj \:=\: 4x,\;\;hyp \:=\: 4x^2+1$


    Pythagorus says: .$\displaystyle (\text{opp})^2 + (\text{adj})^2 \;=\;(\text{hyp})^2$

    So we have: .$\displaystyle (\text{opp})^2 + (4x)^2 \:=\:(4x^2+1)^2 \quad\Rightarrow\quad(\text{opp})^2 + 16x^2 \:=\:16x^4 + 8x^2 + 1$

    . . . . . . . . . .$\displaystyle (\text{opp})^2 \;=\;16x^2 - 8x^2 + 1 \;=\;(4x^2-1)^2 $

    . . . . . . So: . . $\displaystyle \text{opp} \:=\:\pm(4x^2-1)$

    . . . .Hence: . $\displaystyle \tan\theta \;=\;\frac{\pm(4x^2-1)}{4x}$


    Therefore: . $\displaystyle \sec\theta + \tan\theta \;=\;\frac{4x^2+1}{4x} \pm\frac{4x^2-1}{4x} \;=\;\begin{Bmatrix} 2x \\ \\[-4mm]\dfrac{1}{2x}\end{Bmatrix} $

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