Solve for x: 2tan⁻¹(1/3) + sec⁻¹((5√2)/7) = tan⁻¹ x

$2\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{2/3}{1 - 1/9} = \tan^{-1}\frac{3}{4}$. Let $\sec^{-1}\frac{5\sqrt{2}}{7} = \theta$: $\cos\theta = \frac{7}{5\sqrt{2}}$, $\sin\theta = \frac{1}{5\sqrt{2}}$, $\tan\theta = \frac{1}{7}$, so $\theta = \tan^{-1}\frac{1}{7}$. $\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{28}} = \tan^{-1}\frac{25/28}{25/28} = \tan^{-1}1$. Hence $x = 1$.