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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Updated on: 10-Oct-2022

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