The value of tan$\left(\frac{1}{4}{\sin }^{-1}\frac{\sqrt{63}}{8}\right)$ is 

Let ${\sin }^{-1}\frac{\sqrt{63}}{8}=\theta \text{.}$ Then $, \sin \theta =\frac{\sqrt{63}}{8}$ and $\cos \theta =\frac{1}{8}\text{. }$

Now,

$\begin{array}{l}\cos \theta =\frac{1}{8}\Rightarrow 2{\cos }^2\frac{\theta }{2}-1=\frac{1}{8}\Rightarrow {\cos }^2\frac{\theta }{2}=\frac{9}{16}\Rightarrow \cos \frac{\theta }{2}=\frac{3}{4}\\ \therefore \tan \left(\frac{1}{4}{\sin }^{-1}\frac{\sqrt{63}}{8}\right)=\tan \frac{\theta }{4}=\sqrt{\frac{1-\cos \frac{\theta }{2}}{1+\cos \frac{\theta }{2}}}=\sqrt{\frac{1-\frac{3}{4}}{1+\frac{3}{4}}}=\frac{1}{\sqrt{7}}\end{array}$