Question:

Step 1: Substitute $x = \tan\theta$ to simplify the inverse trigonometric expressions. Since $x \geq \frac{4}{3} > 1$, we let $x = \tan\theta$ where $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Step 2: Simplify $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ using the double angle formula. Using the identity $\frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$, we have: $$\sin^{-1}\left(\frac{2x}{1+x^2}\right) = \sin^{-1}(\sin 2\theta)$$ Since $\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, we have $2\theta \in \left(\frac{\pi}{2}, \pi\right)$. In this range, $\sin^{-1}(\sin 2\theta) = \pi - 2\theta$. Step 3: Simplify $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ using the double angle formula. Using the identity $\frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta$, we have: $$\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \cos^{-1}(\cos 2\theta)$$ For $x \geq 0$, this simplifies to $2\theta = 2\tan^{-1}x$. Step 4: Calculate $\alpha$ by substituting the simplified expressions. $$\alpha = \frac{1}{3}(\pi - 2\theta) + \frac{1}{3}(2\theta) = \frac{1}{3}\pi = \frac{\pi}{3}$$ Notice that $\alpha$ is constant regardless of the value of $x$. Step 5: Evaluate $\cos 2\alpha$ and $\sec\alpha$. $$\cos 2\alpha = \cos\frac{2\pi}{3} = -\frac{1}{2}$$ $$\sec\alpha = \sec\frac{\pi}{3} = \frac{1}{\cos\frac{\pi}{3}} = \frac{1}{\frac{1}{2}} = 2$$ Step 6: Calculate the final expression. $$\frac{\cos 2\alpha + \sec\alpha + 3\sqrt{3}}{\sqrt{3}} = \frac{-\frac{1}{2} + 2 + 3\sqrt{3}}{\sqrt{3}}$$ $$= \frac{\frac{3}{2} + 3\sqrt{3}}{\sqrt{3}} = \frac{3}{2\sqrt{3}} + \frac{3\sqrt{3}}{\sqrt{3}}$$ $$= \frac{3}{2\sqrt{3}} + 3 = \frac{3\sqrt{3}}{6} + 3 = \frac{\sqrt{3}}{2} + 3$$ The value of $\frac{\cos 2\alpha + \sec\alpha + 3\sqrt{3}}{\sqrt{3}}$ is $\boxed{\frac{\sqrt{3}}{2} + 3}$, which corresponds to **Option 4**.