Let the expression be \(E = \tan \left[ 2 \tan^{-1} \frac{1}{2} - \frac{\pi}{4} \right]\).
First, simplify the term \(2 \tan^{-1} \frac{1}{2}\). Let \(\theta = \tan^{-1} \frac{1}{2}\), which implies \(\tan \theta = \frac{1}{2}\).
Using the double angle identity for tangent, \(\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}\), we calculate:
\(\tan \left( 2 \tan^{-1} \frac{1}{2} \right) = \frac{2 \times \frac{1}{2}}{1 - \left( \frac{1}{2} \right)^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}\).
The original expression now simplifies to \(\tan \left[ \frac{4}{3} - \frac{\pi}{4} \right]\).
Apply the tangent subtraction formula: \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\).
Here, let \(A\) be the angle such that \(\tan A = \frac{4}{3}\) and \(B = \frac{\pi}{4}\). We know \(\tan B = \tan \frac{\pi}{4} = 1\).
Substitute the values into the subtraction formula:
\(\tan \left[ \frac{4}{3} - \frac{\pi}{4} \right] = \frac{\frac{4}{3} - 1}{1 + \frac{4}{3} \times 1} = \frac{\frac{4 - 3}{3}}{\frac{3 + 4}{3}} = \frac{\frac{1}{3}}{\frac{7}{3}} = \frac{1}{7}\).
Therefore, the value of the expression \(\tan \left[ 2 \tan^{-1} \frac{1}{2} - \frac{\pi}{4} \right]\) is \(\frac{1}{7}\).
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