If \(A + B = 90^\circ\) and \(\tan A = 2\), determine the value of \(\tan A \times \tan B\).
\(1\)
Since \(A + B = 90^\circ\), the two angles are complementary, so \(B = 90^\circ - A\).
Use the complementary-angle identity \(\tan(90^\circ - A) = \cot A\). Therefore \(\tan B = \cot A = \frac{1}{\tan A}\).
Now multiply: \(\tan A \times \tan B = \tan A \times \frac{1}{\tan A} = 1\).
Notice the product does not depend on the actual value of \(\tan A\); the given \(\tan A = 2\) simply confirms the angles are well defined.
Hence, \(\tan A \times \tan B = 1\).
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