Question
Without using trignometric tables, evaluate:
$\frac{\sin16^\circ}{\cos74^\circ}$

Answer

$\frac{\sin16^\circ}{\cos74^\circ}$
$=\frac{\sin(90^\circ-74^\circ)}{\cos74^\circ}$
$=\frac{\cos74^\circ}{\cos74^\circ}$ $[\therefore\ \sin(90-\theta)=\cos\theta]$
$=1$

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