Answer
$sec~(sin^{-1}~(-\frac{1}{5})) = \frac{5~\sqrt{24}}{24}$
Work Step by Step
$\theta = sin^{-1}(-\frac{1}{5})$
$sin~\theta = -\frac{1}{5} = \frac{opposite}{hypotenuse}$
Note that $\theta$ is in quadrant IV. We can find the magnitude of the adjacent side:
$adjacent = \sqrt{5^2-1^2} = \sqrt{24}$
In quadrant IV, $sec~\theta$ is positive. We can find the value of $sec~\theta$:
$sec~\theta = \frac{hypotenuse}{adjacent}$
$sec~\theta = \frac{5}{\sqrt{24}}$
$sec~\theta = \frac{5~\sqrt{24}}{24}$
Therefore, $sec~(sin^{-1}~(-\frac{1}{5})) = \frac{5~\sqrt{24}}{24}$
