Let $y=\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)$.....................(i)
Let $\cos ^{-1} \frac{\sqrt{5}}{3}=\theta \Rightarrow \theta \in\left(0, \frac{\pi}{2}\right)$ and $\cos \theta=\frac{\sqrt{5}}{3}$
$\therefore \quad$ (i) becomes $y=\tan \left(\frac{\theta}{2}\right)$................(ii)
$\because \quad \tan ^2 \frac{\theta}{2}=\frac{1-\cos \theta}{1+\cos \theta}=\frac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}=\frac{3-\sqrt{5}}{3+\sqrt{5}}=\frac{(3-\sqrt{5})^2}{4} \\$
$\tan \frac{\theta}{2}= \pm\left(\frac{3-\sqrt{5}}{2}\right) \\$....................(iii)
$\frac{\theta}{2} \in\left(0, \frac{\pi}{4}\right) \quad \Rightarrow \quad \tan \frac{\theta}{2}>0 \\$
$\therefore \quad \text { from (iii), we get } y=\tan \frac{\theta}{2}=\left(\frac{3-\sqrt{5}}{2}\right)$
