Question

Evaluate :
$sin{18}^{\circ }/cos{72}^{\circ }=$___.

Answer 1

$sin{18}^{\circ }/cos{72}^{\circ }$
$=sin({90}^{\circ }-{72}^{\circ })/cos{72}^{\circ }$
$=cos{72}^{\circ }/cos{72}^{\circ }=1$ (As, sin (${90}^{\circ }$ - A) = cos A)