Answer
$0$
Work Step by Step
We know that $\tan(\theta)=\cot(90^{o}-\theta)$.
Hence,
$\tan(12^{o})-\cot(78^{o})
\\=\cot(90^{o}-12^{o})-\cot(78^{o})
\\=\cot(78^{o})-\cot(78^{o})
\\=0$
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