If $\sec A-\cos A=1$, then determine the value of $\tan^2\frac A2$ This is what I tried
$\sec A=\frac{1}{\cos A}$, so the equation becomes 
$1-\cos^2A=\cos A$
If we solve the above quadratic equation, we the values of $\cos A$ as $\frac{-1\pm \sqrt5}{2}$
Therefore, $\tan\frac A2$ becomes 
$$\sqrt \frac{3-\sqrt 5}{1+\sqrt 5}$$
Squaring that value, the answer remains meaningless
The options are 
A) $\sqrt 5+ 2$
B) $\sqrt 5-2$
C) $2-\sqrt5$
D) $0$
Since the options are not matching, where am I going wrong?

Just notice that
$$
\frac{3-\sqrt 5}{1+\sqrt 5}=\frac{(3-\sqrt 5)(1-\sqrt{5})}{(1+\sqrt 5)(1-\sqrt{5})}=\frac{-4\sqrt{5}+8}{-4}=\sqrt{5}-2.
$$
