# How do you evaluate tan(sec^-1 4)?

Oct 13, 2016

$\tan \left({\sec}^{-} 1 4\right) = \sqrt{15}$

#### Explanation:

$\tan \left({\sec}^{-} 1 4\right)$

The restriction for the range of ${\sec}^{-} 1 x$ is $\left[0 , \pi\right] , y \ne \frac{\pi}{2}$ and since the argument is positive it means that our triangle is in quadrant I with the hypotenuse of 4 and adjacent of 1 and therefore using pythagorean theorem we have the opposite is $\sqrt{15}$.

Therefore, the ratio for tangent from our triangle is
$\tan \theta = \frac{o}{a} = \frac{\sqrt{15}}{1} = \sqrt{15}$