If $ \alpha = 2 tan^{-1} (\sqrt{2} - 1),$

$\beta = 3sin^{-1}\frac{1}{\sqrt{6}}+sin^{-1}(-\frac{1}{2})$ and $γ= cos^{-1} \frac{1}{3}, $ Then,

$ α <γ<  β $

We have,

$ \alpha = 2 tan^{-1} (\sqrt{2} - 1) = 2 tan^{-1} (tan \frac{\pi}{8}) = 2 × \frac{\pi}{8}=\frac{\pi}{4} = cos^{-1} \frac{1}{\sqrt{2}}$

$\beta = 3 sin^{-1}\frac{1}{\sqrt{2}}+sin^{-1}(-\frac{1}{2}) = 3  ×\frac{\pi}{4}-\frac{\pi}{6}=\frac{7\pi}{12}$

and, $ \gamma = cos^{-1}\frac{1}{3}$

Clearly, $\alpha < \beta $

Clearly, $ cos^{-1}x$ is a decreasing function on [-1, 1] and $1/3 < 1/\sqrt{2}.$

$∴ cos^{-1} \frac{1}{3} > cos^{-1} \frac{1}{\sqrt{2}} ⇒ γ > α.$

We observe that

$0 < \gamma < \frac{\pi}{2} $ and $ \frac{\pi}{2} < \beta < \pi ⇒ γ < \beta $

Thus, we have $\alpha < \gamma < \beta  $