1. **Problem:** Find the value of $\tan \left[2 \cos \left(2 \sin^{-1} \frac{1}{2}\right)\right]$. 2. **Step 1: Evaluate the inner inverse sine function.** We know $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$ because $\sin \frac{\pi}{6} = \frac{1}{2}$. 3. **Step 2: Substitute and simplify inside the cosine.** Calculate $2 \sin^{-1} \frac{1}{2} = 2 \times \frac{\pi}{6} = \frac{\pi}{3}$. 4. **Step 3: Find $\cos \left(\frac{\pi}{3}\right)$.** $\cos \frac{\pi}{3} = \frac{1}{2}$. 5. **Step 4: Multiply by 2 inside the tangent.** $2 \cos \left(2 \sin^{-1} \frac{1}{2}\right) = 2 \times \frac{1}{2} = 1$. 6. **Step 5: Calculate $\tan(1)$.** Since 1 is in radians, $\tan(1)$ is the tangent of 1 radian. 7. **Final answer:** $$\tan \left[2 \cos \left(2 \sin^{-1} \frac{1}{2}\right)\right] = \tan(1)$$ This is an exact expression; numerically, $\tan(1) \approx 1.5574$.