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.
 A: $$\frac {7^2}{24^2}=\frac {1+\cos 2\alpha}{1-\cos 2\alpha} \implies$$      $$\implies 7^2(1-\cos \alpha)=24^2(1+\cos \alpha)\implies$$     $$\implies 7^2- 7^2\cos 2\alpha  = 24^2+ 24^2 \cos 2\alpha\implies$$ $$\implies  7^2-24^2= (7^2+24^2)\cos 2\alpha =25^2 \cos 2\alpha\implies$$ $$\implies -527=625\cos 2\alpha .$$
The missing negative sign  on the LHS of the above line is your first error.
Your second error is writing $\cos \frac {\alpha}{2}=\sqrt {\frac {1+\cos \alpha}{2}}\;.$ We have $|\cos \frac {\alpha}{2}|=\sqrt {   \frac {1+\cos \alpha}{2}   }\;.\;$.... If $450^o<\alpha<340^o$ then $225^o<\frac {\alpha}{2}<270^o,$ implying $\cos \frac {\alpha}{2}<0.$
In general if $\cot x=\frac {a}{b}$ then the proportion of $\cos^2 x$ to $\sin^2 x$  is $a^2$ to $b^2$, so let $\cos^2 x=a^2y$ and $\sin^2 x=b^2y$. Since $1=\cos^2 x +\sin^2 x$, we have $y=a^2+b^2$, so $\cos^2 x =\frac {a^2}{a^2+b^2}$ and $\sin^2 x =\frac {b^2}{a^2+b^2}\;$ and therefore  $\;|\cos x|=\frac {|a|}{\sqrt {a^2+b^2}}$ and $|\sin x|=\frac {|b|}{\sqrt {a^2+b^2}}.$
So if $\cot \alpha =\frac {-7}{24}$ and $\cos \alpha<0$ then $\cos \alpha =-\frac {7}{\sqrt {7^2+24^2}}=-\frac {7}{25}.$
A: It's probably easier to use $$\cos^2\alpha = \frac{1}{\sec^2\alpha} = \frac{1}{1+\tan^2\alpha} = \frac{1}{1+\frac{1}{\cot^2\alpha}}$$ to find $\cos\alpha$, this gives $\cos\alpha=\pm\frac7{25}$. The given range for $\alpha$ tells which of these applies. 
I'd suggest you check your calculation of $\cos\alpha$.
A: 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
$$-\frac1{\sqrt2}<\cos\frac\alpha2<0.$$
A: $$\cot\alpha = -\frac{7}{24} \implies \tan\alpha = -\frac{24}{7}$$
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,
$$\cos\left(\frac{\alpha}{2}\right) = -\sqrt{\frac{1 + \cos\alpha}{2}} = -\sqrt{\frac{1 - \frac{7}{25}}{2}} = -\sqrt{\frac{\frac{18}{25}}{2}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$$  
A: $$\cot x =\frac{-7}{24}=\frac{\cos x}{\sin x}$$
$$\frac{49}{576}=\frac{\cos^2 x}{1-\cos^2 x}$$
It gives $$\cos x=\frac{-7}{25}$$
And 
$$\cos \frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}=\frac{3}{5}=0.6 $$ 
Hence (A) is correct 
EDIT:
After concerning with experts , I concluded that $\cos \frac{x}{2} $ is negative and hence (D) is correct
