Answer
$1$
Work Step by Step
We know that $\sin(\theta)=\cos(90^{o}-\theta)$.
Hence,
$\dfrac{\cos(40^{o})}{\sin(50^{o})}
\\=\dfrac{\cos(40^{o})}{\cos(90^{o}-50^{o})}
\\=\dfrac{\cos(40^{o})}{\cos(40^{o})}
\\=1$
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.