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?

Let $t=\tan^2\dfrac A2$. Then $\cos A=\dfrac{1-t}{1+t}$.
$\dfrac{1+t}{1-t}-\dfrac{1-t}{1+t}=1$
$4t=1-t^2$
$(t+2)^2=5$
$t=-2+\sqrt5$
