Correct option is (A) \(\frac 1{\sqrt 7}\)
Let \(\sin^{-1} \left(\frac{\sqrt{63}}8\right) = \theta\)
⇒ \(\sin \theta =\frac{\sqrt{63}}8\)
So,
\(\tan \left(\frac 14 \sin^{-1} \frac{\sqrt{63}}8\right) = \tan \frac \theta 4\)

\(\cos \theta = \frac 18\)
⇒ \(2 \cos^2\frac \theta 2 - 1 = \frac 18\)
⇒ \( \cos^2\frac \theta 2 = \frac 9{16}\)
⇒ \(\cos \frac \theta 2 = \frac 34\)
⇒ \(\frac{1 - \tan^2\frac \theta 4}{1 + \tan^2\frac \theta 4} = \frac 34\)
\(\therefore \tan \frac \theta 4 = \frac 1{\sqrt 7}\)