Question Number 207516 by MATHEMATICSAM last updated on 17/May/24

Commented by A5T last updated on 17/May/24

Answered by A5T last updated on 17/May/24
![3[sin^2 θ+cos^2 θ−2sinθcosθ]^2 +6(1+2sinθcosθ) +4[(sin^2 θ+cos^2 θ)^3 −3sin^2 θcos^2 θ(sin^2 θ+cos^2 θ)] =3[1−2sinθcosθ]^2 +6+6sin2θ+4[1−3sin^2 θcos^2 θ] =3[1+4sin^2 θcos^2 θ−4sinθcosθ]+6+4 −12sin^2 θcos^2 θ+6sin2θ =13−12sinθcosθ+6sin2θ=13](https://www.tinkutara.com/question/Q207517.png)
Commented by Frix last updated on 17/May/24

Commented by A5T last updated on 17/May/24

Commented by Frix last updated on 17/May/24

Answered by Frix last updated on 17/May/24
