If A doesnot belong to the first quadrant, B does not belong to the second quadrant sinA=1161,cosB=−725 then A−B and A+B lie respectively in the quadrants

GivenSin⁡A=1161⇒Alies in 2ndquadrant Cos⁡B=−725⇒B lies in 3rdquadrant ⇒Cos⁡A=−6061,Sin⁡B=−2425 Now,Sin⁡(A−B)=Sin⁡ACos⁡B−cos⁡ASin⁡B =(1161)(−725)−(−6061)(−2425) 0∴A−B lie in 4th quadrant Sin⁡(A+B)=(1161)(−725)+(−6061)(−2425)>0 Cos⁡(A+B)=(−725)(−6061)−(1161)(−2425)>0∴A+B lie in 1st quadrant

If A doesnot belong to the first quadrant, B does not belong to the second quadrant sinA=1161,cosB=−725 then A−B and A+B lie respectively in the quadrants

GivenSin⁡A=1161⇒Alies in 2ndquadrant Cos⁡B=−725⇒B lies in 3rdquadrant ⇒Cos⁡A=−6061,Sin⁡B=−2425 Now,Sin⁡(A−B)=Sin⁡ACos⁡B−cos⁡ASin⁡B =(1161)(−725)−(−6061)(−2425) 0∴A−B lie in 4th quadrant Sin⁡(A+B)=(1161)(−725)+(−6061)(−2425)>0 Cos⁡(A+B)=(−725)(−6061)−(1161)(−2425)>0∴A+B lie in 1st quadrant

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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

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4, 1

2, 3

3, 4

1,2

answer is D.

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Detailed Solution

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