Answer
$\dfrac{4}{\sqrt {15}}$ or, $\dfrac{4\sqrt {15}}{15}$
Work Step by Step
Suppose $\theta =\sin^{-1} (-\dfrac{1}{4})$
This gives: $\sin \theta=-\dfrac{1}{4}$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ x=\sqrt {r^2-y^2}=\sqrt {42^2-(1)^2}=\sqrt 5$
Therefore, we have $\sec \theta =\sec [\sin^{-1} (-\dfrac{1}{4})]=\dfrac{4}{\sqrt {15}}=\dfrac{4\sqrt {15}}{15}$
