Given $\sec\theta=-\frac{13}{12}$ find $\cos{\frac{\theta}{2}}$, where $\frac\pi2<\theta<\pi$.
If the $\sec\theta$ is $-\frac{13}{12}$ then, the $\cos \theta$ is $-\frac{12}{13}$, and the half angle formula tells us that $\cos{\frac{\theta}{2}}$ should be
$$\sqrt{\frac{1+\left(-\frac{12}{13}\right)}{2}}$$
which gives me $\sqrt{\dfrac{1}{26}}$ which rationalizes to $\dfrac{\sqrt{26}}{26}$.
The worksheet off which I'm working lists the answer as $\dfrac{5\sqrt{26}}{26}$.
Can someone explain what I've done wrong here?