# How do you evaluate sec(cos^-1(1/3)) without a calculator?

$\sec \left({\cos}^{-} 1 \left(\frac{1}{3}\right)\right) = 3$
The restriction for the range of arccos x is $\left[0 , \pi\right]$. Since the argument is positive it means our triangle is in quadrant I with adjacent of 1 and hypotenuse 3. Hence using pythagorean theorem we find that the opposite is $2 \sqrt{2}$.
Now let find the ratio for the secant of the angle from the triangle which is $\sec \theta = \frac{h}{a} = \frac{3}{1} = 3$