1. The problem is to verify the equation $$2\tan^{-1}\frac{1}{2} - \tan^{-1}\frac{1}{7} = \frac{\pi}{4}$$. 2. Recall the formula for the tangent of a difference: $$\tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}$$. 3. Let $$A = \tan^{-1}\frac{1}{2}$$ and $$B = \tan^{-1}\frac{1}{7}$$. Then the left side is $$2A - B$$. 4. Use the double angle formula for tangent: $$\tan(2A) = \frac{2 \tan A}{1 - \tan^2 A}$$. 5. Calculate $$\tan(2A)$$: $$\tan(2A) = \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$. 6. Now, the left side is $$2A - B$$, so: $$\tan(2A - B) = \frac{\tan(2A) - \tan B}{1 + \tan(2A) \tan B} = \frac{\frac{4}{3} - \frac{1}{7}}{1 + \frac{4}{3} \cdot \frac{1}{7}} = \frac{\frac{28}{21} - \frac{3}{21}}{1 + \frac{4}{21}} = \frac{\frac{25}{21}}{\frac{25}{21}} = 1$$. 7. Since $$\tan(2A - B) = 1$$, it follows that $$2A - B = \frac{\pi}{4}$$ (principal value). 8. Therefore, the equation is verified. Final answer: $$2\tan^{-1}\frac{1}{2} - \tan^{-1}\frac{1}{7} = \frac{\pi}{4}$$.