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

Question

For any $\theta \in\left({\pi\over4},{\pi\over2}\right)$, the expression $3(\sin\theta-\cos\theta)^4+6(\sin\theta+\cos\theta)^2+4\sin^6\theta$ equals :

(1) $13-4\cos^6\theta$

(2) $13-4\cos^4\theta+2\sin^2\theta\cos^2\theta$

(3) $13-4\cos^2\theta+6\cos^4\theta$

(4) $13-4\cos^2\theta+6\sin^2\theta\cos^2\theta$