Answer
$sec~(2~sin^{-1}~(-\frac{1}{3})) = \frac{9}{7}$
Work Step by Step
Let $\theta = sin^{-1}~(-\frac{1}{3})$
Then $sin~\theta = -\frac{1}{3}$
We can find the value of $sec~2\theta$:
$sec~2\theta = \frac{1}{cos~2\theta}$
$sec~2\theta = \frac{1}{1-2sin^2~\theta}$
$sec~2\theta = \frac{1}{1-2(-\frac{1}{3})^2}$
$sec~2\theta = \frac{1}{1-(\frac{2}{9})}$
$sec~2\theta = \frac{1}{(\frac{7}{9})}$
$sec~2\theta = \frac{9}{7}$
$sec~(2~sin^{-1}~(-\frac{1}{3})) = \frac{9}{7}$
