let's say for a second that that angle is θ, so we can say that
[tex]sin^{-1}\left(\cfrac{\sqrt{7}}{7} \right)=\theta \qquad therefore\qquad tan\left[ sin^{-1}\left(\cfrac{\sqrt{7}}{7} \right) \right]\implies tan(\theta ) \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{this really means}}{sin^{-1}\left(\cfrac{\sqrt{7}}{7} \right)=\theta}\implies \cfrac{\stackrel{opposite}{\sqrt{7}}}{\underset{hypotenuse}{7}}=sin(\theta )[/tex]
let us notice something, the opposite side is positive, that means either the I or II Quadrant, however the sin⁻¹ function has a range constrained to π/2 to -π/2, which excludes the II Quadrant, so that means that θ must be on the I Quadrant, where the adjacent side or cosine or "x" is positive too.
now, let's use the pythagorean theorem to get the adjacent side for θ
[tex]\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=\stackrel{hypotenuse}{7}\\ a=adjacent\\ b=\stackrel{opposite}{\sqrt{7}}\\ \end{cases} \\\\\\ \pm\sqrt{7^2-(\sqrt{7})^2}=a\implies \pm\sqrt{42}=a\implies \stackrel{I~Quadrant}{+\sqrt{42}=a} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]tan(\theta )\implies \cfrac{\stackrel{opposite}{\sqrt{7}}}{\underset{adjacent}{\sqrt{42}}}\implies \cfrac{\sqrt{7}\cdot \sqrt{42}}{\sqrt{42}\cdot \sqrt{42}}\implies \cfrac{\sqrt{294}}{42}\implies \cfrac{7\sqrt{6}}{42}\implies \cfrac{\sqrt{6}}{6}[/tex]
