Correct Answer: $\frac{2+\sqrt{5}}{2 \sqrt{5}}$

Solution :

In $\triangle ABC, \angle B=90^{\circ}$
AB : BC = 1 : 2
⇒ $\frac{\text{AB}}{\text{BC}} = \frac{1}{2}$
⇒ $\frac{\text{AB}}{1} = \frac{\text{BC}}{2}=$ k
⇒ AB = k, BC = 2k
Using the Pythagoras theorem,
AC$^2$ = BC$^2$ + AB$^2$ = (2k)$^2$ + k$^2$ = 5k$^2$
So, AC = $\sqrt{5}$k
By definition of trigonometric ratios,
$\cos A = \frac{\text{AB}}{\text{AC}} = \frac{1}{\sqrt{5}}$
$\tan C = \frac{\text{AB}}{\text{BC}} =  \frac{1}{2}$
⇒ $\cos A+\tan C$ = $ \frac{1}{2}+ \frac{1}{\sqrt{5}}$ = $\frac{\sqrt{5}+2}{2 \sqrt{5}}$
Hence, the correct answer is $\frac{2+\sqrt{5}}{2 \sqrt{5}}$.