Question

Find :
$\frac{{cos18}^o}{{sin72}^o}$

Answer 1

$\frac{\cos {18}^{\circ }}{\sin {72}^{\circ }}$

$=\frac{\cos ({90}^{\circ }-{72}^{\circ })}{\sin {72}^{\circ }}$ since $\cos (90-\theta )=\sin \theta$

$=\frac{\sin {72}^{\circ }}{\sin {72}^{\circ }}=1$