Precalculus (6th Edition) Blitzer

$3$
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, $\dfrac{1}{\cos \theta}=\dfrac{1}{1/3}=3$