Question

Evaluate $\frac{\sin {18}^o}{\cos {72}^o}$

Answer 1

Consider $\sin {18}^0$ which can be rewritten as follows:

$\sin {18}^0=\sin {({90}^0-72}^0)=\cos {72}^0(\because \sin ({90}^0-x)=\cos x)$

Therefore, $\frac{\sin {18}^0}{\cos {72}^0}$ can be evaluated as shown below:

$\frac{\sin {18}^0}{\cos {72}^0}=\frac{\cos {72}^0}{\cos {72}^0}=1$

Hence, $\frac{\sin {18}^0}{\cos {72}^0}=1$.