Question

$\frac{\sin {18}^o}{\cos {72}^o}=$

Answer 1

$\frac{\sin {18}^{\circ }}{\cos {72}^{\circ }}=\frac{\sin ({90}^{\circ }-{72}^{\circ })}{\cos {72}^{\circ }}=\frac{\cos {72}^{\circ }}{\cos {72}^{\circ }}=1$