How do you evaluate ${sec}^{- 1} (\frac{2}{\sqrt{3}})$ $sec^-1( 2/ sqrt3)$ ?

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How do you evaluate ${sec}^{- 1} (\frac{2}{\sqrt{3}})$ $sec^-1( 2/ sqrt3)$ ?

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Answer 1 · 2015-11-25T00:06:58

$\frac{π}{6}$ $pi/6$

Explanation:

Ask yourself: "secant of what angle gives me $\frac{2}{\sqrt{3}}$ $2/sqrt3$ ?"

Since $sec θ = \frac{1}{cos θ}$ $sectheta=1/costheta$ , an easier question to ask would be, "cosine of what angle gives me $\frac{\sqrt{3}}{2}$ $sqrt3/2$ ?"

We know that $cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ $cos(pi/6)=sqrt3/2$ , so $sec (\frac{π}{6}) = \frac{2}{\sqrt{3}}$ $sec(pi/6)=2/sqrt3$ .

While secant is also positive in quadrant four, and there are an infinite amount of coterminal angles where secant is $\frac{2}{\sqrt{3}}$ $2/sqrt3$ , the domain of ${sec}^{- 1} (x)$ $sec^-1(x)$ is restricted from $(0, π)$ $(0,pi)$ , so $\frac{π}{6}$ $pi/6$ is the only valid angle.

Answer 2 · 2015-11-26T03:03:33

$30^{\circ}$ $30^@$ or $330^{\circ}$ $330^@$

Explanation:

${sec}^{- 1} (\frac{2}{\sqrt{3}})$ $sec^-1(2/sqrt(3))$
$sec θ = \frac{2}{\sqrt{3}}$ $sectheta=2/sqrt(3)$
$\frac{1}{cos θ} = \frac{2}{\sqrt{3}}$ $1/costheta=2/sqrt(3)$
$cos θ = \frac{\sqrt{3}}{2}$ $costheta=sqrt(3)/2$
$θ = {cos}^{- 1} (\frac{\sqrt{3}}{2})$ $theta=cos^-1(sqrt(3)/2)$
$θ = 30^{\circ}$ $theta=30^@$

However, since $\frac{2}{\sqrt{3}}$ $2/sqrt(3)$ is positive, there is more than one answer because according to the CAST rule, $cos$ $cos$ is positive in more than one quadrant:

![http://mathonline.wikidot.com/cast-rule]( useruploads.socratic.org )

$cos$ $cos$ is positive in quadrants $1$ $1$ and $4$ $4$ .

To find the other angle that would also give an answer of $\frac{2}{\sqrt{3}}$ $2/sqrt(3)$ , subtract $30^{\circ}$ $30^@$ from $360^{\circ}$ $360^@$ . This ensures that your principal angle is in quadrant $4$ $4$ :

$360^{\circ} - 30^{\circ}$ $360^@-30^@$
$= 330^{\circ}$ $=330^@$

$:.$ , $sec^-1(2/sqrt(3))$ is $30^@$ or $330^@$ .

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How do you evaluate ${sec}^{- 1} (\frac{2}{\sqrt{3}})$ $sec^-1( 2/ sqrt3)$ ?

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