If A doesnot belong to the first quadrant, B does not belong to the second quadrant  $\sin A=\frac{11}{61},\cos B=\frac{-7}{25}$ then  $A-B$ and  $A+B$ lie respectively in the quadrants 

$\begin{array}{rl} & \text{Given}\mathrm{Sin}A=\frac{11}{61}\Rightarrow A\text{lies in} 2^{\text{nd}}\text{quadrant}\\ & \mathrm{Cos}B=\frac{-7}{25}\Rightarrow \text{B lies in }{3}^{\text{rd}}\text{quadrant}\\ & \Rightarrow \mathrm{Cos}A=\frac{-60}{61},\mathrm{Sin}B=\frac{-24}{25}\\ & \text{Now,}\mathrm{Sin}(A-B)=\mathrm{Sin}A\mathrm{Cos}B-\cos A\mathrm{Sin}B\\ & =\left(\frac{11}{61}\right)\left(\frac{-7}{25}\right)-\left(\frac{-60}{61}\right)\left(\frac{-24}{25}\right)<0\\ & \mathrm{Cos}(A-B)=\mathrm{Cos}A\mathrm{Cos}B+\mathrm{Sin}A\mathrm{Sin}B\\ & =\left(\frac{-7}{25}\right)\left(\frac{-60}{61}\right)-\left(\frac{11}{61}\right)\left(\frac{-24}{25}\right)>0\end{array}$
$\therefore \mathrm{A}-\mathrm{B} \text{lie in }{4}^{\text{th}} \text{quadrant}$
$\begin{array}{rl} & \mathrm{Sin}(A+B)=\left(\frac{11}{61}\right)\left(\frac{-7}{25}\right)+\left(\frac{-60}{61}\right)\left(\frac{-24}{25}\right)>0\\ & \mathrm{Cos}(A+B)=\left(\frac{-7}{25}\right)\left(\frac{-60}{61}\right)-\left(\frac{11}{61}\right)\left(\frac{-24}{25}\right)>0\end{array}$
$\therefore \mathrm{A}+\mathrm{B} \text{lie in }{1}^{\text{st}} \text{quadrant}$