Question

Evaluate the following :
$\frac{cos19}{sin71}$

Answer 1

We have $\sin (90-\theta )=\cos \theta$

therefore,

$\sin 71=\sin (90-19)=\cos 19$

To evaluate,

$\frac{\cos 19}{\sin 71}$

$=\frac{\cos 19}{\cos 19}$

$=1$