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

Given:

\( \frac{\sin 18^{\circ}}{\cos 72^{\circ}} \)

To do:

We have to evaluate \( \frac{\sin 18^{\circ}}{\cos 72^{\circ}} \).

Solution:  

We know that,

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

Therefore,

$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}=\frac{\sin 18^{\circ}}{\cos( 90^{\circ}-18^{\circ})}$

$=\frac{\sin 18^{\circ}}{\sin 18^{\circ}}$

$=1$

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

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Updated on: 10-Oct-2022

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