1. **State the problem:** Prove that $$\tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}(2) - \tan^{-1}(3) = \frac{\pi}{4}$$. 2. **Recall the formula for the sum of inverse tangents:** $$\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1 - xy}\right)$$ provided that $$xy < 1$$ or the angles are adjusted accordingly. 3. **Calculate the sum of the first two terms:** Let $$x = \frac{4}{3}$$ and $$y = 2$$. $$\tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}(2) = \tan^{-1}\left(\frac{\frac{4}{3} + 2}{1 - \frac{4}{3} \times 2}\right)$$ Simplify numerator: $$\frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}$$ Simplify denominator: $$1 - \frac{4}{3} \times 2 = 1 - \frac{8}{3} = \frac{3}{3} - \frac{8}{3} = -\frac{5}{3}$$ So, $$\tan^{-1}\left(\frac{10/3}{-5/3}\right) = \tan^{-1}\left(\frac{10}{3} \times \frac{3}{-5}\right) = \tan^{-1}\left(-2\right)$$ 4. **Use the property of inverse tangent:** $$\tan^{-1}(-a) = -\tan^{-1}(a)$$ So, $$\tan^{-1}(-2) = -\tan^{-1}(2)$$ 5. **Rewrite the original expression:** $$\tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}(2) - \tan^{-1}(3) = \tan^{-1}(-2) - \tan^{-1}(3)$$ 6. **Use the formula for difference of inverse tangents:** $$\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right)$$ Let $$a = -2$$ and $$b = 3$$: $$\tan^{-1}(-2) - \tan^{-1}(3) = \tan^{-1}\left(\frac{-2 - 3}{1 + (-2)(3)}\right) = \tan^{-1}\left(\frac{-5}{1 - 6}\right) = \tan^{-1}\left(\frac{-5}{-5}\right) = \tan^{-1}(1)$$ 7. **Evaluate:** $$\tan^{-1}(1) = \frac{\pi}{4}$$ **Final answer:** $$\tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}(2) - \tan^{-1}(3) = \frac{\pi}{4}$$ This completes the proof.