Let [tex]x=\cos^{-1}\dfrac{12}{13}[/tex], so that [tex]\cos x=\dfrac{12}{13}[/tex].
Recall that
[tex]\cos^2x+\sin^2x=1\implies\sin x=\dfrac5{13}[/tex]
where we take the positive root because [tex]\cos\theta[/tex] is only invertible if [tex]0\le\theta<\pi[/tex], and [tex]\cos\theta>0[/tex] only if [tex]0\le\theta<\dfrac\pi2[/tex], which means [tex]\sin\theta>0[/tex].
Now,
[tex]\tan\left(\cos^{-1}\dfrac{12}{13}\right)=\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac5{13}}{\frac{12}{13}}=\dfrac5{12}[/tex]
