Question

Evaluate:$\frac{\sin {18}^0}{\cos {72}^0}$

Answer 1

To evaluate : $\frac{\sin {18}^0}{\cos {72}^0}$

We can write ${18}^0$ as ${18}^0=({90}^0-{72}^0)$

So $\sin {18}^0=\sin ({90}^0-{72}^0)$

⇒ $\sin {18}^0=\cos {72}^0$ [Formula used: $\sin ({90}^0-\mathrm{\theta })=\mathrm{cos\theta }$]

Now, replacing the value of $\sin {18}^0$from the above , we get

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

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