Answer
$1$
Work Step by Step
Recall the co-function identity:
$ \sin (\theta)=\cos(90^{\circ}-\theta)$
Therefore, $\dfrac{\cos (40^{\circ})}{\sin (50^{\circ})}
\\=\dfrac{\cos \ 40^{\circ}}{\cos(90^{\circ}- 50^{\circ})}
\\=\dfrac{\cos \ 40^{\circ}}{\cos \ 40^{\circ}}
\\=1$