Question

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

Answer 1

$\frac{\sin {18}^{\circ }}{\cos {72}^{\circ }}=\frac{sin{18}^0}{cos(90-18)}$
$=\frac{sin{18}^0}{sin{18}^0}=1$