1. **State the problem:** We want to prove that $$\arccos\frac{5}{13} = 2\arctan\frac{2}{3}$$. 2. **Recall the formulas:** - The double angle formula for tangent is $$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$$. - The relationship between cosine and tangent for an angle $$\theta$$ is $$\cos\theta = \frac{1}{\sqrt{1 + \tan^2\theta}}$$ if $$\theta$$ is in the first quadrant. 3. **Set $$\theta = \arctan\frac{2}{3}$$:** - Then $$\tan\theta = \frac{2}{3}$$. 4. **Calculate $$\tan(2\theta)$$ using the double angle formula:** $$\tan(2\theta) = \frac{2 \times \frac{2}{3}}{1 - \left(\frac{2}{3}\right)^2} = \frac{\frac{4}{3}}{1 - \frac{4}{9}} = \frac{\frac{4}{3}}{\frac{5}{9}} = \frac{4}{3} \times \frac{9}{5} = \frac{12}{5}$$. 5. **Find $$\cos(2\theta)$$ using $$\tan(2\theta)$$:** $$\cos(2\theta) = \frac{1}{\sqrt{1 + \tan^2(2\theta)}} = \frac{1}{\sqrt{1 + \left(\frac{12}{5}\right)^2}} = \frac{1}{\sqrt{1 + \frac{144}{25}}} = \frac{1}{\sqrt{\frac{169}{25}}} = \frac{1}{\frac{13}{5}} = \frac{5}{13}$$. 6. **Conclusion:** Since $$\cos(2\theta) = \frac{5}{13}$$ and $$\theta = \arctan\frac{2}{3}$$, it follows that $$2\arctan\frac{2}{3} = \arccos\frac{5}{13}$$. Thus, the identity is proven.