# How do you evaluate sec^-1( 2/ sqrt3)?

Nov 25, 2015

$\frac{\pi}{6}$

#### Explanation:

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

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

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

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

Nov 26, 2015

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

#### Explanation:

${\sec}^{-} 1 \left(\frac{2}{\sqrt{3}}\right)$
$\sec \theta = \frac{2}{\sqrt{3}}$
$\frac{1}{\cos} \theta = \frac{2}{\sqrt{3}}$
$\cos \theta = \frac{\sqrt{3}}{2}$
$\theta = {\cos}^{-} 1 \left(\frac{\sqrt{3}}{2}\right)$
$\theta = {30}^{\circ}$

However, since $\frac{2}{\sqrt{3}}$ is positive, there is more than one answer because according to the CAST rule, $\cos$ is positive in more than one quadrant: $\cos$ is positive in quadrants $1$ and $4$.

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

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

$\therefore$, ${\sec}^{-} 1 \left(\frac{2}{\sqrt{3}}\right)$ is ${30}^{\circ}$ or ${330}^{\circ}$.