Answer
$3$
Work Step by Step
Since, $\csc( \dfrac{\pi}{2}-\theta)=\sec \theta $
The trigonometric ratios are as follows:
$\sin \theta =\dfrac{y}{r} \\ \cos \theta =\dfrac{x}{r} \\ \tan \theta =\dfrac{y}{x}\\ \csc \theta =\dfrac{r}{y} \\ \sec \theta =\dfrac{r}{x} \\ \cot \theta =\dfrac{x}{y}$
where, $ r=\sqrt {x^2+y^2}$
Now, we have
$\csc( \dfrac{\pi}{2}-\theta)=\sec \theta $
or, $\dfrac{1}{\cos \theta}=\dfrac{1}{1/3}=3$
