Question

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

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, $\sin$ (${90}^{\circ }-A)$ = $\cos A$ )