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

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

(C) $13-4 \cos ^{2} \theta+6 \cos ^{4} \theta$

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

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Trigonometric Ratios and Identities

For any

θ \in (\frac{π}{4}, \frac{π}{2})

, the expression

3 (sin θ - cos θ)^{4} + 6 (sin θ + cos θ)^{2} + 4 sin^{6} θ

equals:

Answer: (B)

Expand carefully.

3 (sin θ - cos θ)^{4} = 3 (1 - sin 2 θ)^{2}

and

6 (sin θ + cos θ)^{2} = 6 (1 + sin 2 θ)

; adding these gives

9 + 3 sin^{2} 2 θ

. Writing

sin^{2} 2 θ = 4 sin^{2} θ cos^{2} θ

and combining with the

4 sin^{6} θ

term, the whole expression simplifies to

13 - 4 cos^{6} θ

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option