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Evaluate the following:$ \frac{\cos 19^{\circ}}{\sin 71^{\circ}} $
Given:
\( \frac{\cos 19^{\circ}}{\sin 71^{\circ}} \)
To do:
We have to evaluate \( \frac{\cos 19^{\circ}}{\sin 71^{\circ}} \)
Solution:
We know that,
$sin\ (90^{\circ}- \theta) = cos\ \theta$
Therefore,
$\frac{\cos 19^{\circ}}{\sin 71^{\circ}}=\frac{\cos 19^{\circ}}{\sin (90^{\circ}-19^{\circ})}$
$=\frac{\cos 19^{\circ}}{\cos 19^{\circ}}$
$=1$
Therefore, $\frac{\cos 19^{\circ}}{\sin 71^{\circ}}=1$.
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