Solve for :

$2\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{2/3}{1 - 1/9}\right) = \tan^{-1}\left(\frac{3}{4}\right)$. Let $\theta = \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) \implies \sec\theta = \frac{5\sqrt{2}}{7}$. $\tan\theta = \sqrt{\sec^2\theta - 1} = \sqrt{\frac{50}{49} - 1} = \frac{1}{7} \implies \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) = \tan^{-1}\left(\frac{1}{7}\right)$. Now, $\tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{3/4 + 1/7}{1 - (3/4)(1/7)}\right) = \tan^{-1}\left(\frac{25/28}{25/28}\right) = \tan^{-1}(1)$. Thus $\tan^{-1} x = \tan^{-1}(1) \implies x = 1$.