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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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