Question
Without using trignometric tables, evaluate:
$\frac{\cot38^\circ}{\tan52^\circ}$
$\frac{\cot38^\circ}{\tan52^\circ}$
✓
Answer
$\frac{\cot38^\circ}{\tan52^\circ}$
$=\frac{\cot(90^\circ-52^\circ)}{\tan52^\circ}$
$=\frac{\tan52^\circ}{\tan52^\circ}$ $\big[\therefore\ \tan(90-\theta)=\cot\theta\big]$
$=1$
$=\frac{\cot(90^\circ-52^\circ)}{\tan52^\circ}$
$=\frac{\tan52^\circ}{\tan52^\circ}$ $\big[\therefore\ \tan(90-\theta)=\cot\theta\big]$
$=1$
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