Answer
$0$
Work Step by Step
We know that $\tan(\theta)=\cot(90^{o}-\theta)$.
Hence,
$\tan(42^{o})-\cot(48^{o})\\
=\cot(90^{o}-42^{o})-\cot(48^{o})\\
=\cot(48^{o})-\cot(48^{o})\\
=0$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 11 - Systems of Equations and Inequalities - 11.3 Systems of Linear Equations: Determinants - 11.3 Assess Your Understanding - Page 744: 73
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