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QJEEM21Ma88e3: JEE Main 2021 Mathematics - Class 11 JEEM21Ma88e3 Solution: If 0 < x, y < and cosx + cosy cos(x + y) = , then sinx + cosy…

If 0 < x, y < π\pi and cosx + cosy −- cos(x + y) = 32{3 \over 2} , then sinx + cosy is equal to :

Multiple Choice Options

  • A 1+32{{1 + \sqrt 3 } \over 2}
  • B 12{{1 \over 2}}
  • C 32{{\sqrt 3 } \over 2}
  • D 1−32{{1 - \sqrt 3 } \over 2}

Step-by-Step Verified Solution

2cos⁡(x+y2)cos⁡(x−y2)−[2cos⁡2(x+y2)−1]=322\cos \left( {{{x + y} \over 2}} \right)\cos \left( {{{x - y} \over 2}} \right) - \left[ {2{{\cos }^2}\left( {{{x + y} \over 2}} \right) - 1} \right] = {3 \over 2}

2cos⁡(x+y2)[cos⁡(x−y2)−cos⁡(x+y2)]=122\cos \left( {{{x + y} \over 2}} \right)\left[ {\cos \left( {{{x - y} \over 2}} \right) - \cos \left( {{{x + y} \over 2}} \right)} \right] = {1 \over 2}

2cos⁡(x+y2)[2sin⁡(x2).sin⁡(y2)]=122\cos \left( {{{x + y} \over 2}} \right)\left[ {2\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right)} \right] = {1 \over 2}

cos⁡(x+y2).sin⁡(x2).sin⁡(y2)=18\cos \left( {{{x + y} \over 2}} \right).\sin \left( {{x \over 2}} \right).\sin \left( {{y \over 2}} \right) = {1 \over 8}

Possible when x2=30∘{x \over 2} = 30^\circ & y2=30∘{y \over 2} = 30^\circ

x=y=60∘x = y = 60^\circ

sin⁡x+cos⁡y=32+12=3+12\sin x + \cos y = {{\sqrt 3 } \over 2} + {1 \over 2} = {{\sqrt 3 + 1} \over 2}

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