JEE Main 2019 — Mathematics Trigonometry

$$3{(\mathrm{sin}\theta -\mathrm{cos}\theta )}^{4}+6{(\mathrm{cos}\theta +\mathrm{sin}\theta )}^{2}+4{sin}^{6}\theta$$3(sinθ−cosθ)4+6(cosθ+sinθ)2+4sin6θ

$$=3{({(\mathrm{sin}\theta -\mathrm{cos}\theta )}^{2})}^{2}+6{(cos\theta +sin\theta )}^{2}+4{sin}^{6}\theta$$=3((sinθ−cosθ)2)2+6(cosθ+sinθ)2+4sin6θ

$$=3{({cos}^{2}\theta +{sin}^{2}\theta -2sin\theta cos\theta )}^{2}+6({cos}^{2}\theta +{sin}^{2}\theta +2sin\theta cos\theta )+4{sin}^{6}\theta$$=3(cos2θ+sin2θ−2sinθcosθ)2+6(cos2θ+sin2θ+2sinθcosθ)+4sin6θ

$$=3{(1-2sin\theta cos\theta )}^{2}+6(1+2sin\theta cos\theta )+4{sin}^{6}\theta$$=3(1−2sinθcosθ)2+6(1+2sinθcosθ)+4sin6θ

$$=3(1-4sin\theta cos\theta +4{sin}^{2}\theta {cos}^{2}\theta )+6(1+2sin\theta cos\theta )+4{sin}^{6}\theta$$=3(1−4sinθcosθ+4sin2θcos2θ)+6(1+2sinθcosθ)+4sin6θ

$$=9+12{sin}^{2}\theta {cos}^{2}\theta +4{sin}^{6}\theta$$=9+12sin2θcos2θ+4sin6θ

$$=9+12(1-{cos}^{2}\theta ){cos}^{2}\theta +4{(1-{cos}^{2}\theta )}^{3}$$=9+12(1−cos2θ)cos2θ+4(1−cos2θ)3

$$=9+12{cos}^{2}\theta -12{cos}^{4}\theta +4(1-3{cos}^{2}\theta +3{cos}^{4}\theta -{cos}^{6}\theta )$$=9+12cos2θ−12cos4θ+4(1−3cos2θ+3cos4θ−cos6θ)

$$=13-4{cos}^{6}\theta$$=13−4cos6θ