Find the value of tan⁻¹ x - cot⁻¹ x, if (tan⁻¹ x)² - (cot⁻¹ x)² = (5π²)/8

Let $t=\tan^{-1}x$; then $\cot^{-1}x=\frac\pi2-t$, so $t+\cot^{-1}x=\frac\pi2$. $(\tan^{-1}x)^2-(\cot^{-1}x)^2=(\tan^{-1}x-\cot^{-1}x)(\tan^{-1}x+\cot^{-1}x)=\frac{5\pi^2}{8}$ $(\tan^{-1}x-\cot^{-1}x)\cdot\frac\pi2=\frac{5\pi^2}{8}$ Hence $\tan^{-1}x-\cot^{-1}x=\frac{5\pi}{4}$.