Answer
$\tan{(A+B)} = 2$
$\cot{(A+B)} = \dfrac{1}{2}$
$QI$
Work Step by Step
$\cos{A} = \sqrt{1-(\dfrac{\sqrt{5}}{5})^2} = \dfrac{2\sqrt{5}}{5}$
$\tan{A} = \dfrac{\sin{A}}{\cos{A}} = \dfrac{1}{2}$
$\tan{(A+B)} = \dfrac{\tan{A}+\tan{B}}{1- \tan{A} \tan{B}} = \dfrac{\dfrac{1}{2}+ \dfrac{3}{4}}{1-(\dfrac{1}{2})( \dfrac{3}{4})}$
$\tan{(A+B)} = 2$
$\cot{(A+B)} = \dfrac{1}{\tan{(A+B)}} = \dfrac{1}{2}$
$\because A,B \in QI \hspace{20pt} \tan{(A+B)} >0 \hspace{20pt} \therefore \text{A+B terminates in QI}$
