Answers: 2
![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 )](/tpl/images/2640/5912/9f572.png)
![\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}](/tpl/images/2640/5912/e3895.png)
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Evaluate without a calculator: tan (sin-1 ((√ 7)/7)...
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