Verified Solution • 2025–26 Session moderate

QJEEM19Mafd6c: JEE Main 2019 Mathematics - Class 11 JEEM19Mafd6c Solution: For any , the expression equals :

For any θ∈(π4,π2)\theta \in \left( {{\pi \over 4},{\pi \over 2}} \right) , the expression

3(cos⁡θ−sin⁡θ)43{(\cos \theta - \sin \theta )^4} +6(sin⁡θ+cos⁡θ)2+4sin⁡6θ+ 6{(\sin \theta + \cos \theta )^2} + 4{\sin ^6}\theta

equals :

Multiple Choice Options

  • A 13 – 4 cos2 θ\theta + 6sin2 θ\theta cos2 θ\theta
  • B 13 – 4 cos6 θ\theta
  • C 13 – 4 cos2 θ\theta + 6cos2 θ\theta
  • D 13 – 4 cos4 θ\theta + 2sin2 θ\theta cos2 θ\theta

Step-by-Step Verified Solution

Given,

3(sin θ\theta −- cos θ\theta )4 + 6(sin θ\theta + cos θ\theta )2 + 4sin6 θ\theta

= 3[(sin θ\theta −- cos θ\theta )2]2 + 6 (sin2 θ\theta + cos2 θ\theta + 2sin θ\theta cos θ\theta ) + 4sin6 θ\theta

= 3[sin2 θ\theta + cos2 θ\theta −- 2sin θ\theta cos θ\theta ]2 + 6(1 + sin2 θ\theta ) + 4sin6 θ\theta

= 3(1 −- sin2 θ\theta )2 + 6(1 + sin2 θ\theta ) + 4sin6 θ\theta

= 3 (1 −- 2 sin2 θ\theta + sin22 θ\theta ) + 6 + 6sin2 θ\theta + 4sin6 θ\theta

= 3 −- 6sin2 θ\theta + 3sin22 θ\theta + 6 + 6sin2 θ\theta + 4sin6 θ\theta

= 9 + 3sin22 θ\theta + 4 sin6 θ\theta

= 9 + 3(2sin θ\theta cos θ\theta )2 + 4(1 −- cos2 θ\theta )3

= 9 + 12sin2 θ\theta cos2 θ\theta + 4 (1 −- cos6 θ\theta −- 3cos2 θ\theta + 3cos4 θ\theta )

= 13 + 12 (1 −- cos2 θ\theta −- 4cos6 θ\theta −- 12cos θ\theta + 12 cos4 θ\theta

= 13 + 12 cos2 θ\theta −- 12 cos4 θ\theta −- 4cos6 θ\theta −- 12 cos2 θ\theta + 12 cos4 θ\theta

= 13 −- 4 cos6 θ\theta

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