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For any θ∈,the expression 3(sinθ – cosθ)4 + 6(sinθ + cosθ)2 + 4sin6θ equals :a)13 – 4 cos6θb)13 – 4 cos4θ+ 2 sin2θcos2θc)13 – 4 cos2θ + 6 cos4θd)13 – 4 cos2θ + 6 sin2θcos2θCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about For any θ∈,the expression 3(sinθ – cosθ)4 + 6(sinθ + cosθ)2 + 4sin6θ equals :a)13 – 4 cos6θb)13 – 4 cos4θ+ 2 sin2θcos2θc)13 – 4 cos2θ + 6 cos4θd)13 – 4 cos2θ + 6 sin2θcos2θCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For any θ∈,the expression 3(sinθ – cosθ)4 + 6(sinθ + cosθ)2 + 4sin6θ equals :a)13 – 4 cos6θb)13 – 4 cos4θ+ 2 sin2θcos2θc)13 – 4 cos2θ + 6 cos4θd)13 – 4 cos2θ + 6 sin2θcos2θCorrect answer is option 'A'. Can you explain this answer?.

Solutions for For any θ∈,the expression 3(sinθ – cosθ)4 + 6(sinθ + cosθ)2 + 4sin6θ equals :a)13 – 4 cos6θb)13 – 4 cos4θ+ 2 sin2θcos2θc)13 – 4 cos2θ + 6 cos4θd)13 – 4 cos2θ + 6 sin2θcos2θCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

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