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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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4,1
Given,
$\sin A=\frac{11}{61} A$ lies in 2 nd quadrant
$\cos B=\frac{-7}{25} B$ lies in 3rd quadrant
$\cos A=\frac{-60}{61}$ and $\sin B=\frac{-24}{25}$
$\because \sin (A-B)=\sin A \cos B-\cos A \sin B$
$=\left(\frac{11}{61}\right)\left(\frac{-7}{25}\right)-\left(\frac{-60}{61}\right)\left(\frac{-24}{25}\right)$
$\sin (A-B) < 0$
$\cos (A-B)=\cos A \cos B+\sin A \sin B$
$=\left(\frac{-7}{25}\right)\left(\frac{-60}{61}\right)+\left(\frac{11}{61}\right)\left(\frac{-24}{25}\right)$
$\cos (A-B)>0$
$\because A-B$ lie in 4 th quadrant
$\sin (A+B)=\left(\frac{11}{61}\right)\left(\frac{-7}{25}\right)+\left(\frac{-60}{61}\right)\left(\frac{-24}{25}\right)$
$\sin (A+B)>0$
$\cos (A+B)=\left(\frac{-7}{25}\right)\left(\frac{-60}{61}\right)-\left(\frac{11}{61}\right)\left(\frac{-24}{25}\right)$
$\cos (A+B)>0$
$\therefore(A+B)$ lies in Ist quadrant
$\sin A=\frac{11}{61} A$ lies in 2 nd quadrant
$\cos B=\frac{-7}{25} B$ lies in 3rd quadrant
$\cos A=\frac{-60}{61}$ and $\sin B=\frac{-24}{25}$
$\because \sin (A-B)=\sin A \cos B-\cos A \sin B$
$=\left(\frac{11}{61}\right)\left(\frac{-7}{25}\right)-\left(\frac{-60}{61}\right)\left(\frac{-24}{25}\right)$
$\sin (A-B) < 0$
$\cos (A-B)=\cos A \cos B+\sin A \sin B$
$=\left(\frac{-7}{25}\right)\left(\frac{-60}{61}\right)+\left(\frac{11}{61}\right)\left(\frac{-24}{25}\right)$
$\cos (A-B)>0$
$\because A-B$ lie in 4 th quadrant
$\sin (A+B)=\left(\frac{11}{61}\right)\left(\frac{-7}{25}\right)+\left(\frac{-60}{61}\right)\left(\frac{-24}{25}\right)$
$\sin (A+B)>0$
$\cos (A+B)=\left(\frac{-7}{25}\right)\left(\frac{-60}{61}\right)-\left(\frac{11}{61}\right)\left(\frac{-24}{25}\right)$
$\cos (A+B)>0$
$\therefore(A+B)$ lies in Ist quadrant
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