If A does not belong to the first quadrant,B | vedclass

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(D) Given,$\sin A = \frac{11}{61}$. Since $A$ is not in the first quadrant and $\sin A > 0$,$A$ must lie in the second quadrant. Thus,$\cos A = -\sqrt

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$4, 1$

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(D) Given,$\sin A = \frac{11}{61}$. Since $A$ is not in the first quadrant and $\sin A > 0$,$A$ must lie in the second quadrant. Thus,$\cos A = -\sqrt{1 - (\frac{11}{61})^2} = -\frac{60}{61}$.Given,$\cos B = \frac{-7}{25}$. Since $B$ is not in the second quadrant and $\cos B For $A-B$:$\sin(A-B) = \sin A \cos B - \cos A \sin B = (\frac{11}{61})(\frac{-7}{25}) - (\frac{-60}{61})(\frac{-24}{25}) = \frac{-77 - 1440}{1525} = \frac{-1517}{1525} $\cos(A-B) = \cos A \cos B + \sin A \sin B = (\frac{-60}{61})(\frac{-7}{25}) + (\frac{11}{61})(\frac{-24}{25}) = \frac{420 - 264}{1525} = \frac{156}{1525} > 0$.Since $\sin(A-B)  0$,$A-B$ lies in the fourth quadrant.For $A+B$:$\sin(A+B) = \sin A \cos B + \cos A \sin B = (\frac{11}{61})(\frac{-7}{25}) + (\frac{-60}{61})(\frac{-24}{25}) = \frac{-77 + 1440}{1525} = \frac{1363}{1525} > 0$.$\cos(A+B) = \cos A \cos B - \sin A \sin B = (\frac{-60}{61})(\frac{-7}{25}) - (\frac{11}{61})(\frac{-24}{25}) = \frac{420 + 264}{1525} = \frac{684}{1525} > 0$.Since $\sin(A+B) > 0$ and $\cos(A+B) > 0$,$A+B$ lies in the first quadrant.

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

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