Given the hyperbolic metric $d{s}^{2}=\frac{d{x}^{2}+d{y}^{2}}{{x}^{2}}$ on the half plane $x>0$, find the length of the arc of the circle ${x}^{2}+{y}^{2}=1$ from $(\mathrm{cos}\alpha ,\mathrm{sin}\alpha )$ to $(\mathrm{cos}\beta ,\mathrm{sin}\beta )$

I found that $d{s}^{2}={\displaystyle \frac{d{\theta}^{2}}{{\mathrm{cos}}^{2}\theta}}$ but when I try to plug in $\pi /3,-\pi /3$, which should give me the arc length of $2\pi /3$,

I get $4\pi /3=\sqrt{{\displaystyle \frac{{(\pi /3-(-\pi /3))}^{2}}{co{s}^{2}(\pi /3)}}}$

I feel like I'm making a simple mistake but I cant place it

I found that $d{s}^{2}={\displaystyle \frac{d{\theta}^{2}}{{\mathrm{cos}}^{2}\theta}}$ but when I try to plug in $\pi /3,-\pi /3$, which should give me the arc length of $2\pi /3$,

I get $4\pi /3=\sqrt{{\displaystyle \frac{{(\pi /3-(-\pi /3))}^{2}}{co{s}^{2}(\pi /3)}}}$

I feel like I'm making a simple mistake but I cant place it