For any θ ∊ (\pi/4,\pi/2) , the expression 3 (\sin\theta-\cos\theta)^4 + 6 (\sin\theta+\cos\theta)^2 + 4 \sin^6 \theta equals : [9-Jan-2019 Shift 1]

Correct Answer is : 13 - 4cos6θ

13 - 4cos^4 \theta + 2 \sin^2\thetacos^2\theta

13 - 4cos^2 \theta + 6 \cos^4\theta

13 - 4cos^2 \theta + + \sin^2\thetacos^2\theta

JEE Main 9 Jan 2019 Shift 1 Solved Paper

Section: Mathematics

Question : 55 of 90

Marks: +1, -0

For any θ ∊ (

) , the expression 3 (sinθ−cosθ)4 + 6 (sinθ+cosθ)2 + 4 sin6θ equals :

[9-Jan-2019 Shift 1]

Go to Question:

Prev Question

Next Question