Question
Evaluate the following:
$\frac{\tan10^\circ}{\cot80^\circ}$

Answer

$\frac{\tan10^\circ}{\cot80^\circ}\Rightarrow\frac{\tan(90^\circ-80^\circ)}{\cot80^\circ}=\frac{\cot80^\circ}{\cot80^\circ}$ $\big[\because \sin(90^\circ-\theta)=\cot\theta\big]$
$=1$

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