Evaluate tan(cos ^-1 3/5) and assume that all angles are in Quadrant I.

[tex]\bf cos^{-1}\left( \frac{3}{5} \right)=\theta \quad \stackrel{\textit{this simply means}}{\implies }\quad cos(\theta )=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}[/tex]

and with that, let us find the opposite side, keeping in mind that we're in the I Quadrant, and thus the opposite side is positive, just like adjacent as well,

[tex]\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-3^2}=b\implies \pm\sqrt{16}=b\implies \pm 4=b\implies \stackrel{I~Quadrant}{+4=b}\\\\ -------------------------------\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{4}}{\stackrel{adjacent}{3}}[/tex]