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 ______.

Let us denote

$$\theta = \cos^{-1}\!\Bigl(\sqrt{\frac{2}{2+\pi^{2}}}\Bigr).$$

Since $$0 \lt \sqrt{\dfrac{2}{2+\pi^{2}}} \lt 1$$, the principal value $$\theta$$ lies in $$\bigl(0,\tfrac{\pi}{2}\bigr).$$ Consequently

$$\cos\theta = \sqrt{\frac{2}{2+\pi^{2}}} \quad\Longrightarrow\quad \sin\theta = \sqrt{1-\cos^{2}\theta} = \sqrt{\frac{\pi^{2}}{2+\pi^{2}}} = \frac{\pi}{\sqrt{2+\pi^{2}}}.$$

1. First term

$$\frac{3}{2}\cos^{-1}\!\Bigl(\sqrt{\frac{2}{2+\pi^{2}}}\Bigr) = \frac{3}{2}\theta.$$

2. Second term Calculate the argument:

$$2\sqrt{2}\pi=\bigl(2\sin\theta\cos\theta\bigr)\!\bigl(2+\pi^{2}\bigr),$$ so

$$\frac{2\sqrt{2}\pi}{2+\pi^{2}} = 2\sin\theta\cos\theta = \sin(2\theta).$$

The principal range of $$\sin^{-1}x$$ is $$[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$$. Because $$\theta\in(0,\tfrac{\pi}{2})$$, we have $$2\theta\in\bigl(\tfrac{\pi}{2},\pi\bigr)$$, so $$\sin^{-1}\!\bigl(\sin(2\theta)\bigr)=\pi-2\theta.$$

Hence

$$\frac14\sin^{-1}\!\Bigl(\frac{2\sqrt{2}\pi}{2+\pi^{2}}\Bigr) =\frac14\bigl(\pi-2\theta\bigr) =\frac{\pi}{4}-\frac{\theta}{2}.$$

3. Third term Using $$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\pi}{\sqrt{2}},$$ we observe

$$\frac{\sqrt{2}}{\pi}=\cot\theta=\tan\!\bigl(\tfrac{\pi}{2}-\theta\bigr).$$

Since $$\tan^{-1}x$$ returns principal values in $$(-\tfrac{\pi}{2},\tfrac{\pi}{2})$$, we obtain

$$\tan^{-1}\!\Bigl(\frac{\sqrt{2}}{\pi}\Bigr)=\frac{\pi}{2}-\theta.$$

4. Adding the three terms

$$\begin{aligned} \text{Required value}&=\frac{3}{2}\theta +\Bigl(\frac{\pi}{4}-\frac{\theta}{2}\Bigr) +\Bigl(\frac{\pi}{2}-\theta\Bigr)\\[4pt] &=\Bigl(\frac{3}{2}\theta-\frac{\theta}{2}-\theta\Bigr) +\Bigl(\frac{\pi}{4}+\frac{\pi}{2}\Bigr)\\[4pt] &=0+\frac{3\pi}{4}=\frac{3\pi}{4}. \end{aligned}$$

Numerically, $$\dfrac{3\pi}{4}\approx2.35619$$ rad, which may be rounded as $$2.35$$ or $$2.36$$.

Answer : 2.35 | 2.36 (either value is accepted).