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For any θ ∈ ( π 4, π 2), the expression 3( sinθ - cosθ )^4+6( sinθ + cosθ )^2+4 sin^6θ equals:

JEE Main 2019 — Mathematics Trigonometry

2019mcqmedium

For any θ∈(π4,π2),\theta \in (\frac{\pi }{4},\frac{\pi }{2}),θ∈(4π​,2π​), the expression 3(sinθ−cosθ)4+6(sinθ+cosθ)2+4sin6θ3{(\mathrm{sin}\theta -\mathrm{cos}\theta )}^{4}+6{(\mathrm{sin}\theta +\mathrm{cos}\theta )}^{2}+4 {sin}^{6}\theta3(sinθ−cosθ)4+6(sinθ+cosθ)2+4sin6θ equals:

Official previous-year question

Held on 9 Jan 2019 · Verified 6 Jul 2026.

Options

  1. A

    13−4cos2θ+6cos4θ13-4{\mathrm{cos}}^{2}\theta +6{\mathrm{cos}}^{4}\theta13−4cos2θ+6cos4θ

  2. B

    13−4cos2θ+6sin2θcos2θ13-4{\mathrm{cos}}^{2}\theta +6{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta13−4cos2θ+6sin2θcos2θ

  3. C

    13−4cos6θ13-4{\mathrm{cos}}^{6}\theta13−4cos6θ

  4. D

    13−4cos4θ+2sin2θcos2θ13-4{\mathrm{cos}}^{4}\theta +2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta13−4cos4θ+2sin2θcos2θ

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