Mathematics · 2019
JEE Main · 9 January 2019, Shift 1 · Q88
For any θ ∈((π)/4, (π)/2), the expression 3( sin θ- cos θ)^4+6( sin θ+ cos θ)^2+4 sin^6 θ equals:
For any $\displaystyle \theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, the expression
$$3(\sin \theta-\cos \theta)^4+6(\sin \theta+\cos \theta)^2+4 \sin ^6 \theta
$$
equals :
Official answer
(4)
$\displaystyle 13-4 \cos ^6 \theta$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.