Question

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

Answer 1

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