Considering only the principal values of the inverse trigonometric functions, the value of $\frac{3}{2}\cos^{-1}\sqrt{\frac{2}{2+\pi^{2}}} + \frac{1}{4}\sin^{-1}\frac{2\sqrt{2}\pi}{2+\pi^{2}} + \tan^{-1}\frac{\sqrt{2}}{\pi}$ is ____.

$\text{Let } \alpha = \tan^{-1}\dfrac{\sqrt{2}}{\pi}$ $\text{To find: } \dfrac{3}{2}\cos^{-1}\sqrt{\dfrac{2}{2+\pi^2}} + \dfrac{1}{4}\sin^{-1}\dfrac{2\sqrt{2}\,\pi}{2+\pi^2} + \alpha$ $\text{Plan: } \tan\alpha = \tfrac{\sqrt2}{\pi} \Rightarrow \sin\alpha,\ \cos\alpha \Rightarrow \text{each term} \to \alpha$ $\tan\alpha = \dfrac{\sqrt{2}}{\pi} \Rightarrow \sin\alpha = \dfrac{\sqrt2}{\sqrt{2+\pi^2}},\quad \cos\alpha = \dfrac{\pi}{\sqrt{2+\pi^2}}$ $\sqrt{\dfrac{2}{2+\pi^2}} = \sin\alpha \Rightarrow \cos^{-1}\sqrt{\dfrac{2}{2+\pi^2}} = \cos^{-1}(\sin\alpha)$ $= \cos^{-1}\!\left(\cos\left(\dfrac{\pi}{2}-\alpha\right)\right) = \dfrac{\pi}{2}-\alpha$ $0<\alpha<\dfrac{\pi}{2} \Rightarrow \dfrac{\pi}{2}-\alpha \in \left(0,\dfrac{\pi}{2}\right) \subset [0,\pi]$ $\therefore\ \cos^{-1}(\sin\alpha) = \dfrac{\pi}{2}-\alpha\ \checkmark$ $\dfrac{2\sqrt2\,\pi}{2+\pi^2} = 2\cdot\dfrac{\sqrt2}{\sqrt{2+\pi^2}}\cdot\dfrac{\pi}{\sqrt{2+\pi^2}} = 2\sin\alpha\cos\alpha = \sin 2\alpha$ $\Rightarrow \sin^{-1}\dfrac{2\sqrt2\,\pi}{2+\pi^2} = \sin^{-1}(\sin 2\alpha)$ $\alpha\approx 0.42 \Rightarrow 2\alpha \approx 0.84 \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ $\therefore\ \sin^{-1}(\sin 2\alpha) = 2\alpha\ \checkmark$ $E = \dfrac{3}{2}\left(\dfrac{\pi}{2}-\alpha\right) + \dfrac{1}{4}(2\alpha) + \alpha$ $= \dfrac{3\pi}{4} - \dfrac{3\alpha}{2} + \dfrac{\alpha}{2} + \alpha = \dfrac{3\pi}{4}$ $\boxed{\dfrac{3\pi}{4}}$