<s>[INST] 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.
 [/INST] $$\cot\alpha = -\frac{7}{24} \implies \tan\alpha = -\frac{24}{7}$$
Since $1 + \tan^2\alpha = \sec^2\alpha$,
$$\sec^2\alpha = 1 + \left(-\frac{24}{7}\right)^2 = 1 + \frac{576}{49} = \frac{625}{49} \implies |\sec\alpha| = \frac{25}{7}$$
Observe that $450^\circ < \alpha < 540^\circ \implies \cos\alpha < 0$.  Hence, 
$$\sec\alpha = -\frac{25}{7} \implies \cos\alpha = -\frac{7}{25}$$
Since $450^\circ < \alpha < 540^\circ$, $225^\circ < \frac{\alpha}{2} < 270^\circ$.  Thus, 
$\cos(\frac{\alpha}{2}) < 0$.  Therefore,
$$\cos\left(\frac{\alpha}{2}\right) = -\sqrt{\frac{1 + \cos\alpha}{2}} = -\sqrt{\frac{1 - \frac{7}{25}}{2}} = -\sqrt{\frac{\frac{18}{25}}{2}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$$  
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