Cosecants of half angles 
The tangent of an angle is $2.4$. Find the cosecant of half the angle.

My tries:
As $\tan A=2.4=\dfrac{12}{5}\implies \sin A=\pm\dfrac{12}{13}=\dfrac{1}{\csc A}$
Also:
$\sin \frac{A}{2}+\cos \frac{A}{2}=\pm\sqrt{1+\sin A}=\pm\dfrac{5}{{\sqrt{13}}}$ , considering $\sin A=\dfrac{12}{13}$
$\sin \frac{A}{2}-\cos \frac{A}{2}=\pm\sqrt{1-\sin A}=\pm\dfrac{1}{{\sqrt{13}}}$, considering $\sin A=\dfrac{13}{13}$
Adding them gives: $\sin \frac{A}{2}=\pm\dfrac{3}{\sqrt{13}}=\dfrac{1}{\csc{\frac A2}}$
I did same by considering $\sin A=-\dfrac{12}{13}$ then also it gave the same result.
But answer provided by the author is $\dfrac{\pm\sqrt{13}}{2}$ and $\dfrac{\pm\sqrt{13}}{3}$.
So what did I miss? please help.

You have
$$
\tan^2A=\frac{144}{25}
$$
so
$$
\frac{1}{\cos^2A}=1+\tan^2A=\frac{169}{25}
$$
and therefore
$$
\cos A=\pm\frac{5}{13}
$$
Hence
$$
\sin\frac{A}{2}=\pm\sqrt{\frac{1-\cos A}{2}}
\qquad
\csc\frac{A}{2}=\pm\sqrt{\frac{2}{1-\cos A}}
$$
Plug in the values.
