Question:
For any $\theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, the expression
$3(\sin \theta-\cos \theta)^{4}+6(\sin \theta+\cos \theta)^{2}+4 \sin ^{6} \theta$
Correct Option: 1
Solution:
We have,
$3(\sin \theta-\cos \theta)^{4}+6(\sin \theta+\cos \theta)^{2}+4 \sin ^{6} \theta$
$=3(1-\sin 2 \theta)^{2}+6(1+\sin 2 \theta)+4 \sin ^{6} \theta$
$=3\left(1-2 \sin 2 \theta+\sin ^{2} 2 \theta\right)+6+6 \sin 2 \theta+4 \sin ^{6} \theta$
$=9+12 \sin ^{2} \theta \cdot \cos ^{2} \theta+4\left(1-\cos ^{2} \theta\right)^{3}$
$=13-4 \cos ^{6} \theta$
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