Question

Evaluate :
$\frac{sin{18}^{\circ }}{cos{72}^{\circ }}$_
[1 Mark]

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) = cos A)
[1 Mark]