Question

Question 1 (i)
Evaluate :
$\frac{sin{18}^{\circ }}{cos{72}^{\circ }}$

Answer 1

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

$=\frac{sin({90}^{\circ }-{72}^{\circ })}{cos{72}^{\circ }}$
As, $\left(sin\right({90}^{\circ })-A)=cosA$

So,
$\Rightarrow \frac{cos{72}^{\circ }}{cos{72}^{\circ }}=1$