If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:

If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$

I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1+\cos2\alpha}{1-\cos2\alpha}}\\ \frac{49}{576}&=\frac{1+\cos2\alpha}{1-\cos2\alpha}\\ 625\cos2\alpha&=527\\ 2\cos^2\alpha-1&=\frac{527}{625}\\ \cos\alpha&=-\frac{24}{25}, \end{align}$$ therefore, $$\begin{align} \cos\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{24}{25}}{2}}\\ &=\sqrt{\frac{1}{50}}\\ &=\frac{\sqrt{2}}{10}. \end{align}$$ But there is not such an answer:

A) $0.6$
B) $\frac{4}{5}$
C) $-\frac{4}{5}$
D) $-0.6$
E) $0.96$

I have checked the evaluating process several times. While I believe that my answer is correct and there is a mistake in the choices, I want to hear from you.

Hint:
$$\cot\alpha=\frac{1-\tan^2\frac\alpha2}{2\tan\frac\alpha2}=\frac{1-t^2}{2t}$$ or
$$t^2+2\cot\alpha\ t-1=0,$$
or
$$t=-\cot\alpha\pm\sqrt{1-\cot^2\alpha}=-\frac34,\frac43.$$
Then $$\cos\frac\alpha2=\pm\frac1{\sqrt{1+t^2}}=\pm\frac1{\sqrt{2-2\cot\alpha\ t}}=\pm\frac45,\pm\frac35.$$
Then given range of $\alpha$ tells you that
$$-\frac1{\sqrt2}<\cos\frac\alpha2<0.$$
